Newton’s Laws, Part 7

Please refer to the previous posts in this series if this post is to make any sense.

Believe it or not, I made a small error in the equations from the last post. I used kilometers instead of meters in part of the equation for the force due to gravity. The two equations for the thrust phase and the coasting phase are:\[ \begin{array}{l} {{F}{=}\hspace{0.33em}{2}{,}{800}{,}{000}\hspace{0.33em}{-}\hspace{0.33em}{(}{16000}\hspace{0.33em}{-}\hspace{0.33em}{57}{.}{2222}{t}{)}\hspace{0.33em}\times\hspace{0.33em}{9}{.}{8}{\left({\frac{6400000}{{6400000}\hspace{0.33em}{+}\hspace{0.33em}{x}{(}{t}{)}}}\right)}^{2}}\\ {{=}\hspace{0.33em}{(}{16000}\hspace{0.33em}{-}\hspace{0.33em}{57}{.}{2222}{t}{)}\hspace{0.33em}\times\hspace{0.33em}{a}} \end{array} \]

and \[ \begin{array}{c} {{F}{=}{-}{9}{.}{8}{\left({\frac{6400000}{{6400000}\hspace{0.33em}{+}\hspace{0.33em}{x}{(}{t}{)}}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{a}} \end{array} \]

Now as mentioned before, these are rather difficult to solve. However, first class rocket scientists rarely solve equations by hand. They resort to numerical methods to solve them. Numerical methods means that they enter the equations into a program and then let the computer solve them.

I have done that and have plotted the results. Below is a plot of the distance the rocket is after t seconds:

Looks like we sent the rocket into space with quite a kick! At 270 seconds, the thrust stops and the rocket keeps going with no sign of slowing down. Its velocity when the thrust stops is greater than the escape velocity at that height. Escape velocity is the speed an object needs to break free of the earth’s gravity. The rocket never falls back to earth.

I also plotted the velocity of the rocket. See how the velocity remains relatively constant after thrust cutoff because gravity is too weak to slow it down:

So that’s all I will say about Newton’s laws, at least for this series of posts. Next week, what’s the deal with this clock?

Newton’s Laws, Part 6

Now let’s use a bigger rocket and remove two of the assumptions we made with the model rocket. If you have not read the previous posts in this series, I suggest that you do before starting cold on this one.

As I said before, the assumption that the rocket’s mass stays the same throughout its flight is not realistic for larger rockets. The fuel is the majority of the mass of a rocket and that is spent as the rocket ascends. The assumption that gravity is constant throughout a rocket flight is also not realistic as gravity does get weaker the higher the rocket goes.

Let’s assume a hypothetical rocket that has a fuelled mass of 16,000 kg. The unfuelled rocket (the rocket structure and payload) weigh 550 kg. We will continue to assume a constant thrust (again not realistic) of 2,800,000 N that lasts 270 seconds. As the fuel burns, let’s assume a constant rate of fuel being ejected. The mass of the rocket, as time t increases is:

m(t) = 16000 – 57.2222 t

this makes sense since at t = 0, the mass of the rocket is 16000 kg and after 270 seconds, the mass is 550 kg. Note that this mass equation is only valid during the thrust phase of the rocket. As before, we have to deal with two separate phases of the rocket’s flight because the equations to be solved are different in each phase.

The change in gravity requires a little explanation. The acceleration due to gravity is due to the attraction between two bodies (you and the earth). The acceleration due to gravity follows what is called the inverse square law. This means that the acceleration due to gravity decreases by the square of the distance between them.

Now the distance between you and the center of the earth is about 6400 km. We use the center of large objects from which to measure distance. As our rocket travels up, this distance is getting greater so gravity is becoming weaker. If x(t) is the distance from the launchpad after t seconds, then g(t), the acceleration due to gravity at t seconds is:

\[
{g}{(}{t}{)}\hspace{0.33em}{=}\hspace{0.33em}{9}{.}{8}{\left({\frac{6400}{{6400}\hspace{0.33em}{+}\hspace{0.33em}{x}{(}{t}{)}}}\right)}^{2}
\]

So during the burn phase, there are two forces acting in the rocket: gravity and the rocket engine. We will follow the same convention as before, anything pointing down is negative, and up is positive. So for phase 1, Newton’s second law, F = ma is

\[
\begin{array}{l}
{{F}{=}\hspace{0.33em}{2}{,}{800}{,}{000}\hspace{0.33em}{-}\hspace{0.33em}{(}{16000}\hspace{0.33em}{-}\hspace{0.33em}{57}{.}{2222}{t}{)}\hspace{0.33em}\times\hspace{0.33em}{9}{.}{8}{\left({\frac{6400}{{6400}\hspace{0.33em}{+}\hspace{0.33em}{x}{(}{t}{)}}}\right)}^{2}}\\
{{=}\hspace{0.33em}{(}{16000}\hspace{0.33em}{-}\hspace{0.33em}{57}{.}{2222}{t}{)}\hspace{0.33em}\times\hspace{0.33em}{a}}
\end{array}
\]

As you can see, removing those two assumptions really complicates the equation. Things get a little simpler for phase 2, after the engines burn out because the mass is now constant at 550 kg and this can be divided out from both sides of the equation:

\[
\begin{array}{c}
{{F}{=}{-}{550}\hspace{0.33em}\times\hspace{0.33em}{9}{.}{8}{\left({\frac{6400}{{6400}\hspace{0.33em}{+}\hspace{0.33em}{x}{(}{t}{)}}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{550}\hspace{0.33em}\times\hspace{0.33em}{a}}\\
{\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{-}{9}{.}{8}{\left({\frac{6400}{{6400}\hspace{0.33em}{+}\hspace{0.33em}{x}{(}{t}{)}}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{a}}
\end{array}
\]

These equations from both phases are rather nasty ones. To get technical, these are non-linear second order differential equations. You can see why rocket scientists make the big bucks.

So this post is long enough so I will say more about these equations next time.

Newton’s Laws, Part 5

Please read the prior posts in this series if you have not been following along. I ended last post with two equations, each relating to a different phase of our model rockets flight: powered and coasting. Let’s look at the powered phase (phase 1).

For this phase, Newton’s second law (F = ma) reduced down to:

5.08 = 0.4a

From this equation, a first class rocket scientist can use calculus to find the velocity and the distance from the launch pad at any time in seconds after launch. These equations are:

v(t) = 12.7t, x(t) = 6.35t²

where v(t) is the velocity t seconds after launch and x(t) is the distance from the launch pad t seconds after launch. Now I’ve introduced what is called functional notation. Instead of saying the velocity or distance after 3 seconds, I can just say v(3) or x(3). Maths is full of shorthand notations.

These two equations assumes that we start the clock at 0 seconds and that velocity, acceleration, and distance are all 0 at 0 seconds. Now remember, these equations are only valid for the first 3.3 seconds of flight (see previous post) because the engine stops burning at 3.3 seconds.

So how fast is our rocket going at 3.3 seconds? Well, just replace t with 3.3 in the velocity equation and calculate it:

v(3.3) = 12.7×3.3 = 41.91 m/s

To give you a perspective of how fast this is, this is equivalent to almost 151 km/h. How high is the rocket at engine burnout? Let’s replace t with 3.3 in the distance equation and calculate it:

x(3.3) = 6.35×(3.3)² = 69.15 m

Is this the highest the rocket goes, a measly 69 meters? Well remember, at burnout, the rocket is going up very fast. It will take gravity a while to turn that around. Enter the phase 2 equations.

From my last post, Newton’s second law for phase 2 is:

a = -9.8

Again, using calculus, our friend, the first class rocket scientist generates the two equations (applicable only to phase 2):

v(t) = -9.8t + 74.25, x(t) = -4.9t² + 74.25t -122.514

These equations are a bit more complex because they have to take into account that at 3.3 seconds, the velocity is 41.91 m/s and the rocket is 69.15 m high.

So how high does our rocket go? At the peak of its travels, the velocity goes from positive (going up) to negative (going down). That is, it passes through 0. So in order to find the highest that our rocket goes, we need to find when the velocity equals 0. So we use our velocity equation and set it equal to o, then solve for the time that makes that happen:

v(t) = -9.8t + 74.25 = 0

-9.8t = -74.25

t = -74.25/-9.8 = 7.58 seconds

So now we use the distance equation and replace t with 7.58:

x(7.58) = -4.9×(7.58)² + 74.25×7.58 -122.514 = 158.76 m

So now remember that we are not using a parachute. So the next two questions to ask is when does it hit the ground and how fast is it going when it does.

When the rocket hits the ground, its distance is 0. So now we use the distance equation, set it equal to 0 and find the value of t to make that happen:

x(t) = -4.9t² + 74.25t -122.514 = 0

Now you can solve this using the quadratic formula which I have covered in a previous post. Using this formula, you get two answers: 1.884s and 13.269s. The first answer is not greater than 3.3. These phase 2 equations are only valid for t greater than 3.3 seconds. So we can reject that answer and choose 13.269 seconds. So the total flight time is a bit over 13 seconds.

Now how fast does it hit the ground? Put the time 13.269 into the velocity equation to get:

v(13.269) = -9.8×13.269 + 74.25 = -55.79 m/s

The velocity is negative because it is going down. So the rocket is going its fastest when it hits the ground, not when the engine burns out. 55.79 m/s is equivalent to 200.84 km/h. What are the odds that we can launch this rocket again?

In my next post, let’s do the same problem but use a bigger rocket (since our model rocket is now one with the earth).

Newton’s Laws, Part 4

Well congratulations! If you get through (and understand) this and the last three posts on Newton’s laws, consider yourself a rocket scientist third class.

In my high school days, my friend Byron and I formed a two-member rocket club. We would pool our money and get and build model rockets. We also got solid propellant cartridges that were inserted into the rocket. We would put the rocket on a launch pad (a wooden block) and ignite the propellant, either with a fuse or electrically with a small coil of wire that was heated with an electric current. The rocket would ascend very fast, then a small charge at the end of the propellant cartridge would push out the nose cone and the folded parachute inside. Then the fun part was to try and retrieve the rocket, especially on windy days!

So today, I am going the find the equation of motion and the velocity equation of these rockets using Newton’s second law, F = ma.

We are going to make several assumptions. I will relax some of these later, but seeing that you are only a third class rocket scientist, let’s start out making things as simple as possible.

Assumptions:

  1. The force exerted by the cartridge is constant from ignition to the time it cuts off.
  2. The mass of the rocket is constant throughout its flight.
  3. The rocket is going straight up and down. (We will not be using a parachute.)
  4. The flight time is short so we will not take into account the rotation of the earth.
  5. The force due to gravity is constant throughout the flight.
  6. Air drag is ignored.

A first class rocket scientist would not be making these assumptions and would include the effects of these into the equations. Assumption 1 is not valid for real rockets and the varying force would have to be accounted for.

Assumption 2 would make the equations very inaccurate in reality. Depending on the type of propellant, the rocket fuel can be from 83% to 96% of the total mass of the rocket. As the fuel is burned, the mass of the rocket is getting less and less, and as you can see from F = ma, this means that the acceleration must increase if m×a is to continue equalling the constant force.

Assumption 3 is made so that the velocity, acceleration, and position are measured along the same line. For big rockets, measurements of these three things are measured with something called vectors which provide a direction as well as a value.

Assumption 4 is made because the earth does rotate and this rotation adds to the speed, not in the vertical direction but in a sideways direction.

Assumption 5 is OK for our model rocket because it will not go that high, but the force of gravity does lessen with altitude and this must be taken into account for larger rockets.

Assumption 6 allows us to keep the equations simple (as do all of them).

Now let’s look at an actual model rocket. These numbers are realistic numbers that are possible for model rockets.

Suppose we launch a model rocket that weighs 400 g. The engine provides 9 N of force for 3.3 s. Let’s answer the following questions:

  1. How high does the rocket go?
  2. What is the maximum speed of the rocket?
  3. What is the total flight time?

Before I set this up, let’s agree on what’s positive and what’s negative. Let’s agree that up is positive and down is negative. So the force due to the rocket engine is positive because it is pushing the rocket up and the force due to gravity is negative because it is pulling the rocket down.

Now there are two phases of the rocket’s flight: the first phase is when the engine is burning and the second phase begins when the engine stops. Let’s look at F = ma during the first phase.

Phase 1 – engine is burning: There are two forces acting on the rocket, the one due to the engine and the other due to gravity. In part 2 of this series of posts, I explained how to find the force due to gravity. The gravity force points down so it is negative. The other force, of course, is due to the rocket engine and it is positive. So putting this into F = ma:

9 N – (0.4 kg)(9.8 m/s²) = 0.4 kg × a

I use 0.4 kg instead of 400 g because we are using SI units (explained in last post). So simplifying this equation and removing the units gives:

5.08 = 0.4a . This is the starting equation for the first phase.

Phase 2 – engine stops: Now let’s look at the second phase. After the engine cuts off, the only force acting on the rocket is gravity. So the only acceleration the rocket is experiencing is only due to gravity. So F = ma becomes:

– (0.4 kg)(9.8 m/s²) = 0.4 kg × a

But you can see from this that a = -9.8 m/s² to make this a true equation. This simple equation is the starting equation for the second phase.

This post is now longer than I thought it would be, so I will continue with this in my next post.

Newton’s Laws, Part 3

What I would like to eventually get to, is to develop the equation of motion of a rocket. An equation of motion is just an equation that calculates an object’s position given a time. I did this without a lot of detail, in my Springy Thingy posts back in February. For this set of posts, I would like to add a bit more development.

Let’s go back to our basic equation that describes motion: F = ma. Let’s look at the a (acceleration) part. Acceleration is a rate of change of velocity. If a car goes from 0 to 100 km/s in 10 seconds, its acceleration is the difference in velocity divided by the time interval:

\[
{a}\hspace{0.33em}{=}\hspace{0.33em}\frac{{100}\hspace{0.33em}{\mathrm{k}}{\mathrm{m}}{/}{\mathrm{s}}}{{10}\hspace{0.33em}{\mathrm{s}}}\hspace{0.33em}{=}\hspace{0.33em}{10}\hspace{0.33em}\frac{\mathrm{k}\mathrm{m}/\mathrm{s}}{\mathrm{s}}\hspace{0.33em}{=}\hspace{0.33em}{10}\hspace{0.33em}{\mathrm{k}\mathrm{m}/\mathrm{s}}^{2}
\]

This means the car increases its velocity 10 km/s each second. So acceleration is the rate that velocity changes per unit of time. However, velocity is also a rate of change measurement. Velocity is the rate of change of position (or distance) per unit of time. An equation of motion is finding the position of an object given a time.

So acceleration is the rate of change of velocity which is the rate of change of position, and it’s position that we want. How do we get there? This is where calculus comes in.

Calculus essentially deals with rate of change equations. It can find the rate of change of position, that is velocity, given a position equation. It can also go backwards and find a position equation, given a velocity equation. In our case, it can take the rate of change of velocity (acceleration) and find the velocity equation and then take the velocity equation and find the position equation, that is the equation of motion.

This ability to take a simple equation like F = ma, and from it, describe the motion of objects is one of the many reasons I love maths.

With this as a background, next time, let’s launch a rocket, then cut off it’s motors and answer questions like:

How high is the rocket when the motors cut off?

How fast is the rocket going when the rockets cut off?

How high does the rocket go?

When will the rocket hit the ground (or will it hit the ground)?

Newton’s Laws, Part 2

So I left off last time with the equation form of Newton’s second law: F = ma. It is important to use consistent units for the three quantities, the force F, the mass m, and the acceleration a.

Now in the USA, they use English units where the unit for acceleration is ft/sec² (feet per second squared), the unit of mass is something called a slug, and the unit for force is lbf (pound-force). We will not be using these units.

In the civilised world (I’m not biased), we use SI units. SI comes from the French Système international  which means the international system of units. For our equation, these units are kg (kilogram) for mass, m/sec² (meters/second squared) for acceleration, and the combination of these units on the right side of F = ma gives kg×m/sec² as the unit of force. This unit is given a special name in SI units, the newton, N, in honour of guess who. So 1 N is the force required to accelerate 1 kg of mass, 1 m/sec². 1 N is about the force an average size apple exerts on your hand.

Now I said I would also explain the difference between mass and weight. Weight is a force exerted by an object due to gravity. An objects weight changes when measured on different planets or moons. Its mass however, is an intrinsic property and remains unchanged regardless of where the object is. Mass is the amount of stuff that makes up the object.

Now the confusion between these two things arises because we commonly use weight, say in kilograms, to mean force. We feel the weight of a 1 kg object in our hands. But this unit is really a kilogram-force (kgf). A kilogram in SI units is a unit of mass, not weight. But fortunately, an object on earth that exerts a force of 1 kgf due to gravity, is defined as having a mass of 1 kg, so it is easy to interchange these units on earth. But elsewhere, the object will exert a different force (kgf) due to gravity but it will still be an object of 1 kg mass.

The kilogram-force unit is not an SI unit – we just naturally use it in everyday conversation. As seen before, the newton (N) is the SI unit of force. So what is the weight of a 1 kg mass in SI units? On earth, when you drop something from a height, its velocity starts as 0 m/s but its speed increases as time goes on. This is why you can jump from a half a meter height without injuring yourself but will do considerable damage if you jump from a 50 meter height. Increasing speed means acceleration. So there is an acceleration due to gravity on earth as well as other planets. On earth, the acceleration due to gravity is 9.8 m/sec². When an object is dropped on earth, its speed increases by 9.8 m/sec every second.

So if we want to calculate the weight of a 1 kg object, we use Newtons second law F = ma where we use the acceleration due to gravity for a:

F = 1 kg × 9.8 m/sec² = 9.8 N

If you want to know your weight in newtons, just take your mass in kg (which equals your weight in kgf), and multiply by 9.8. Your weight in newtons is almost 10 times your weight in kgf, which explains why people prefer to use kgf.

Newton’s Laws, Part 1

As my interest in maths grew out of seeing it applied, I thought I should start writing posts on its applications. Physics applications are almost always maths related and you can’t start a conversation about physics without starting with Newton’s Laws.

Sir Isaac Newton is famous for his work in physics and maths. I find it amazing that his accomplishments occurred in the 1600’s. One of his most famous works, some would say his most famous, was his Principia Mathematica Philosophiae Naturalis (The Mathematical Principles of Natural Philosophy). In his day, the term natural philosophy meant science. In this publication, Newton set out his three laws of motion:

  1. Every object in a state of uniform motion will remain in that state of motion unless an external force acts on it.
  2. Force equals mass times acceleration.
  3. For every action there is an equal and opposite reaction.

The first law is sometimes called the law of inertia. You experience inertia every day when you try to push an object or stop an object from moving. You have to apply a force to start an object moving or stop one from moving. The second law explains why a greater force is needed to stop a moving car than a baby stroller. The third law explains why rockets work, what happens when you release a ballon before it’s tied, and why a gun or rifle has a recoil.

This series of posts will be mainly about Newton’s second law. This law, in equation form, is

F = ma , where F is force, m is mass, and a is acceleration.

Before we work with this equation using numbers, let’s see what this equation means.

If you try to stop a rolling car, you are trying to decrease its velocity from a certain value to zero. In other words, you are trying to decelerate it. Deceleration is negative acceleration, and according to Newtons second law, because the mass of the car is rather large, a large force is required to stop it from rolling. A baby stroller going the same speed requires less force because its mass, m, is much smaller than a car’s mass.

Even though this equation is very simple, there are entire books dedicated to this equation. The rest of my posts could easily be about this equation alone, but I will try to keep it down to just a few.

In my next post, I will talk about the units that we will use for the three things that make up this equation. I will also talk about the difference between mass and weight.

What Goes Up …

I was a mediocre student in maths before I started at a university. But in university, I discovered the amazing things that maths can do. Maths can be used to describe all sorts of physical processes and this can be used to control those processes. Things like the cruise control in your car or the flight controls of an aircraft or spacecraft need a mathematical model of the thing being controlled. One of the first examples used to introduce students to modelling (that is mathematically describing something), is the simple act of throwing a ball into the air.

Now to start modelling this process from scratch requires calculus which I haven’t covered yet. But I will give you the final result and we will work with that.

If a tennis ball is hit straight up in the air, there are two main forces acting on it after it leaves the racquet: gravity and drag from the air. Though the air drag can be modelled as well, it complicates the model so it is usually assumed to be negligible when introducing this to students. The effect of gravity is to reduce the initial speed given to the ball by 9.8 m/s every second. So if the ball has an initial speed of 60 m/s, after 1 second its speed is 60 – 9.8 = 50.2 m/s. After the next second, its speed is 50.2 – 9.8 = 40.4 m/s, and so on. By the way, I am using metric units here. The same thing can be done with American units where the effect of gravity reduces the speed of the ball by 32 ft/s every second. But as most of the world uses metric, we’ll stay with that.

So let’s stick with the initial velocity of the ball as 60 m/s when it leaves the racket. Now the first thing to do is agree on a coordinate system. It’s natural to agree that up is positive and down is negative. We’ll also agree that time starts at 0 as soon as the ball leaves the racket and that the height in meters at the point where the ball leaves the racquet is also 0. We’ll call this the ground level. So using calculus with the force of gravity and the initial speed of the ball (ignoring air drag), we can get three equations that describe the motion of the ball: one equation for its acceleration, one for its velocity, and one for its height from the ground. With a = acceleration, v = velocity, h = height, and  t = time in seconds, these equations are:

\[
\begin{array}{l}
{{a}\hspace{0.33em}{=}\hspace{0.33em}{-}{9}{.}{8}}\\
{{v}\hspace{0.33em}{=}\hspace{0.33em}{-}{9}{.}{8}{t}\hspace{0.33em}{+}\hspace{0.33em}{60}}\\
{{h}\hspace{0.33em}{=}\hspace{0.33em}{-}{4}{.}{9}{t}^{2}\hspace{0.33em}{+}\hspace{0.33em}{60}{t}}
\end{array}
\]

With these equations, we can answer the following questions:

  1. When will the ball hit the ground?
  2. How fast is the ball travelling when it is at its maximum height?
  3. How high does the ball go?
  4. How fast is the ball going when it hits the ground?

Before I answer these questions, let’s look at the graph of the height equation. Now before, we talked about x and y values on a graph. But for physical processes, we can use different letters that are more meaningful. Instead of x, we will use t for time, and instead of y, we will use h for height. The graph of this equation is below:

The curve is an upside down parabola. The graph goes on forever below the t-axis, but I’m only show the part of the graph that makes physical sense.

The graph and the equation make sense at t = 0 seconds as h is 0 on the graph at t = 0, and if you let t = 0 in the equation, you also get h = 0.

So you can see that the ball goes up, reaches a maximum height, then falls back to the ground. To answer question 1, from the graph, it looks like the ball hits the ground a little over 12 seconds because that is where the graph shows h = 0 again. To find the exact value, we must set the height equation equal to zero, and find the times when h = 0. This will require us to factor the equation and use the null factor law (explained in a previous post):

\[
\begin{array}{l}
{{h}\hspace{0.33em}{=}\hspace{0.33em}{-}{4}{.}{9}{t}^{2}\hspace{0.33em}{+}\hspace{0.33em}{60}{t}\hspace{0.33em}{=}\hspace{0.33em}{t}{(}{-}{4}{.}{9}{t}\hspace{0.33em}{+}\hspace{0.33em}{60}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}}\\
{\Longrightarrow\hspace{0.33em}{-}{4}{.}{9}{t}\hspace{0.33em}{+}\hspace{0.33em}{60}\hspace{0.33em}{=}\hspace{0.33em}{0}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{-}{4}{.}{9}{t}\hspace{0.33em}{=}\hspace{0.33em}{-}{60}}\\
{\Longrightarrow{t}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{60}}{{-}{4}{.}{9}}\hspace{0.33em}{=}\hspace{0.33em}{12}{.}{245}\hspace{0.33em}{\mathrm{seconds}}}
\end{array}
\]

So the ball travels for 12.245 seconds before it hits the ground (that’s a powerful tennis player!). Notice that there is also another solution, t = 0, which is when the ball is initially hit.

Question 2 can be answered by the physics of the problem and this answer will help us answer question 3. As explained, the ball has an initial velocity of 60 m/s, but is continually slowing down due to gravity. When it reaches its maximum height, the ball reverses direction then goes back down. At the maximum height, the velocity is 0 because velocity is positive going up and negative going down, so it must be 0 right when the ball reverses direction.

Question 3 can be answered using the answer to question 2. If we set the velocity equation to zero, that will give us the time that the ball is at maximum height. We then use this time in the height equation to find the height at that time:

\[
\begin{array}{l}
{{v}\hspace{0.33em}{=}\hspace{0.33em}{-}{9}{.}{8}{t}\hspace{0.33em}{+}\hspace{0.33em}{60}\hspace{0.33em}{=}\hspace{0.33em}{0}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{-}{9}{.}{8}{t}\hspace{0.33em}{=}\hspace{0.33em}{-}{60}}\\
{\Longrightarrow\hspace{0.33em}{t}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{60}}{{-}{9}{.}{8}}\hspace{0.33em}{=}\hspace{0.33em}{6}{.}{122}\hspace{0.33em}{\mathrm{seconds}}}\\
{{h}\hspace{0.33em}{=}\hspace{0.33em}{-}{4}{.}{9}{(}{6}{.}{122}{)}^{2}\hspace{0.33em}{+}\hspace{0.33em}{60}{(}{6}{.}{122}{)}\hspace{0.33em}{=}\hspace{0.33em}{183}{.}{67}\hspace{0.33em}{\mathrm{meters}}}
\end{array}
\]

We also could have solved this by noticing that a parabola is symmetric and the maximum height would occur halfway between 0 and 12.245 seconds. This also would have given us t = 6.122 seconds to use in the height equation.

The last question is answered by using the time that the ball hits the ground in the velocity equation:

\[
{v}\hspace{0.33em}{=}\hspace{0.33em}{-}{9}{.}{8}{(}{12}{.}{245}{)}\hspace{0.33em}{+}\hspace{0.33em}{60}\hspace{0.33em}{=}\hspace{0.33em}{-}{60}\hspace{0.33em}{\mathrm{m}}{/}{\mathrm{s}}
\]

Notice that the velocity is the same as when the ball was first hit except it is negative since it is now going down instead of up.

Isn’t it amazing how we can find out all sorts of things about throwing a ball without actually doing it!

Latitude Aptitude

How did the early sailors determine their latitude position without GPS? That is the topic of today’s post.

Now first, a little background. The earth’s axis is tilted with respect to its orbit about the sun. The angle of this tilt is approximately 23.5°. This causes the northern and southern hemispheres to get more sun in summer and less in winter, which is the reason for seasons to exist. The tilted axis also causes our days to be shorter in the winter and longer in the summer. There are two times during the year when the days and nights are equal in length. The times are called the vernal and autumnal equinoxes. In the northern hemisphere, these  equinoxes occur on the first days of spring and autumn. Here in Australia in the southern hemisphere, we elected to call the start of spring on the 1st of September and the fall on the 1st of March, about 21 days short of the respective equinox. Perhaps this is because it is easier to remember. The main point here is that twice a year, at an equinox, the days and nights are equal.

At any time of the year other than an equinox, the highest height of the sun around noon is affected by the tilt of the earth’s axis. But at an equinox, the earth is in a neutral position where the axis tilt does not affect the highest sun height. At the equator (0° latitude), the sun would be directly overhead and a vertical stick in the ground would cast no shadow. As you go up or down in latitude, the highest sun height goes down and a vertical stick would cast the shortest shadow when the sun is at its highest. The below graphic shows the earth at an equinox with the sun at its maximum height. If a vertical stick is placed in the ground at your location, the sun’s rays would make an angle with it that is the same as your latitude angle.

Below is a blow-up of the vertical stick. You can see from the above picture that at the equator. The sun would be directly overhead at noon and there would be no shadow. At the poles, the sun would be at the horizon and the shadow would be very long (technically infinite). But in between, a measurable shadow would be made. 

Now you could measure the angle directly with a sextant, but I hardly know what a sextant is. let alone use one. But I am good at maths and I have a good calculator. The shadow, stick, and the line from the top of the stick to the shadow end forms a right triangle. If you remember the post on trig functions, the tangent of an angle is the length of the opposite side divided by the length of the adjacent side. We want to measure the angle 𝝀, so the adjacent side is the stick and the opposite side is the shadow: 

\[\tan\mathit{\lambda}\hspace{0.33em}{=}\hspace{0.33em}\frac{{\mathrm{length}}\hspace{0.33em}{\mathrm{of}}\hspace{0.33em}{\mathrm{stick}}}{{\mathrm{length}}\hspace{0.33em}{\mathrm{of}}\hspace{0.33em}{\mathrm{shadow}}}\]

On your calculator, if you have the trig functions, you would also have keys labelled “arctan” or “tan-1“. These keys mean “what is the angle that has what you entered as its tangent”. So if you enter the results of the division and then hit this key (making sure that your calculator is in “degrees” mode), you will get your latitude.

Now this method will not tell you if the latitude is positive (North) or negative (South). But if you are so lost that you don’t even know what hemisphere you are in, finding your latitude is probably the least of your troubles.

Also, waiting for noon to find your latitude is not too bad, but waiting for an equinox is fairly restrictive. Fortunately, our early sailors had tables to correct the angle found depending on the time of the year.