Question

A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s2 until its engines stop at an altitude of 150 m. Solve the following questions:
(a) What can you say about the motion of the rocket after its engines stop?
(b) What is the maximum height reached by the rocket?
(c) How long after liftoff does the rocket reach its maximum height?
(d) How long is the rocket in the air?

Answer 1

In the question where as a model rocket is launched straight upward with an initial speed of 50.0m/s and it accelerate with a constant acceleration of 2m/s^2 until it engines stop at an altitude od 150. So the following are the answer to your question:
a.The motion, the acceleration was equals to Zero
b.the maximum height is 150m

Answer 2

Answer:

Part a)

the rocket will decelerate due to gravity and its speed will start decreasing after that

Part b)

$H = 308 m$

Part c)

$T = 8.5 s$

Part d)

$T = 16.4 s$

Explanation:

Part a)

When Engine stops then the acceleration of the rocket is only due to gravity

So the rocket will decelerate due to gravity and its speed will start decreasing after that

Part b)

final speed of the rocket just before its engine is stopped is given as

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 50^2 = 2(2)(150)$

$v_f = 55.67 m/s$

now after this rocket will move under gravity

$v_f^2 - v_i^2 = 2a d$

$0 - 55.67^2 = 2(-9.81)d$

$d = 158 m$

so maximum height from ground is given as

$H = 150 + 158 = 308 m$

Part c)

Time taken by the rocket to reach 150 m height is given as

$v_f = v_i + at$

$55.67 = 50 + 2t_1$

$t_1 = 2.835 s$

Now after this time to reach the maximum height is given as

$0 - 55.67 = -9.81(t_2)$

$t_2 = 5.67 s$

total time is given as

$T = 2.835 + 5.67 = 8.5 s$

Part d)

Time taken by the rocket to drop back on the surface is given as

$y = \frac{1}{2}gt^2$

$308 = \frac{1}{2}(9.81)t^2$

$t = 7.92 s$

so total time in air

$T = 7.92 + 8.5 = 16.4 s$