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).