You have designed a rocket to be used to sample the local atmosphere for pollution. It is fired vertically with a constant upward acceleration of 15 m/s2. After 21 s, the engine shuts off and the rocket continues rising (in freefall) for a while. (Neglect any effects due to air resistance.) The rocket eventually stops rising and then falls back to the ground. You want to get a sample of air that is 11 km above the ground.
(a) What is the highest point your rocket reaches?
a. h1 = 0.5a*t^2 = 7.5*21^2 = 3308 m
V = a*t = 15 * 21 = 315 m/s.
hmax = h1 + (V^2-Vo^2)/2g =
3308 + (0-315^2)/-19.6 = 8371 m.
To find the highest point the rocket reaches, we need to determine the time it takes for the rocket to reach its highest point.
We know that the rocket's initial upward velocity is zero, and it has a constant upward acceleration of 15 m/s^2. We can use the equation of motion:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the rocket's final velocity at the highest point will be zero, we can rewrite the equation as:
0 = 0 + 15t
Simplifying, we find:
t = 0 seconds
This means that the rocket will reach its highest point instantaneously when the engine shuts off. Now, let's find the rocket's position at this point.
We can use the equation:
s = ut + (1/2)at^2
where s is the displacement (distance traveled), u is the initial velocity, a is the acceleration, and t is the time.
At the highest point, the rocket's displacement will be its maximum height. We know that the initial velocity is zero and the acceleration is 15 m/s^2. Plugging in these values, we get:
s = 0(0) + (1/2)(15)(0^2)
s = 0 meters
Therefore, the highest point the rocket reaches is at ground level (0 meters).