Question

A model rocket blasts off and moves upward with an acceleration of 12 m/ss until it reaches a height of 29 m, at which point its engine shuts off and it continues its flight in free fall.

Answer 1

upward acceleration of rocket is given as

$a = 12 m/s^2$

$d = 29 m$

now the speed of rocket is given as

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

$v_f^2 - 0 = 2*12*29$

$v_f = 26.4 m/s$

Here we can use kinematics again as after this the rocket starts free fall so its acceleration is a = 9.8 m/s^2

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

$0 - 26.4^2 = 2*(-9.8)*d$

$d = 35.5 m$

so the maximum height achieved by the rocket will be

$H = 35.5 + 29 = 64.51 m$

now the time of flight of rocket will be given as

first it will reach to height h = 29 m

$v_f - v_i = 2 a d$

$26.4 - 0 = 2*12 * t$

$t_1 = 1.1 s$

now after this it will move under free fall so time to reach the ground is given by

$d = v_i*t + \frac{1}{2} at^2$

$-29 = 26.4*t - \frac{1}{2}*9.8*t^2$

$4.9t^2 - 26.4 t - 29 = 0$

by solving above equation

$t = 6.32 s$

So total time of motion is

$T = t_1 + t_2 = 6.32 + 1.1 = 7.42 s$