Question

A model rocket blasts off and moves upward with an acceleration of 12 m/s2 until it reaches a height of 26 m. at that height, its engine shuts off, and it continues its flight in free fall. (a) what is the maximum height attained by the rocket? (b) what is the speed of the rocket just before it hits the ground? (c) what is the total duration (ins) of the rocket's flight?

Answer 1

initial acceleration of rocket is given as

a = 12 m/s^2

h = 26 m

now we can use kinematics to find its speed

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

$v_f^2 - 0 = 2 * 12*26$

$v_f = 24.98 m/s$

now after this it will be under free fall

so now again using kinematics

$v_f = 0$

at maximum height

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

$0 - 24.98^2 = 2 * (-9.8)* h$

$h = 31.8 m$

total height from the ground = 31.8 + 26 = 57.8 m

Part b)

now after reaching highest height it will fall to ground

So in order to find the speed we can use kinematics again

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

$v_f^2 - 0 = 2*9.8*57.8$

$v_f = 33.67 m/s$

Part c)

first rocket accelerate to reach height 26 meter and speed becomes 24.98 m/s

now we have

$v_f - v_i = a t$

$24.98 - 0 = 12*t_1$

$t_1 = 2.1 s$

after this it will reach to highest point and final speed becomes zero

$v_f - v_i = at$

$0 - 24.98 = -9.8 * t$

$t_2 = 2.55 s$

now from this it will fall back to ground and reach to final speed 33.67 m/s

now we have

$v_f - v_i = at$

$33.67 - 0 = 9.8 * t$

$t_3 = 3.44 s$

so total time is given as

t = 3.44 + 2.55 + 2.1 = 8.1 s