Question

A model rocket is launched straight upward with an initial speed of 57.0 m/s. It accelerates with a constant upward acceleration of 3.00 m/s2 until its engines stop at an altitude of 140 m. (a) What can you say about the motion of the rocket after its engines stop? This answer has not been graded yet. (b) What is the maximum height reached by the rocket? m (c) How long after liftoff does the rocket reach its maximum height? s (d) How long is the rocket in the air? s

Answer 1

Answer:

(a) The motion of rocket is in upward direction and reaches to maximum height then the rocket starts falling freely.

(b) The maximum height attained by the rocket from the ground is 348.65 m

(c) The time taken by the rocket to maximum height after lift off is 8.85 s.

(d) The total time taken by the rocket in air is 17.3 second.

Explanation:

u = 57 m/s, a = 3 m/s^2, h = 140 m

let the rocket attains a velocity v after covering 140 m and it takes t time to reach upto 140 m.

Use III equation of motion

V^2 = u^2 + 2a h

v^2 = 57^2 + 2 x 3 x 140

v = 63.95 m/s

Now use I equation of motion

v = u + at

t = (63.95 - 57) / 3 = 2.32 s

(a) The motion of rocket is in upward direction and reaches to maximum height then the rocket starts falling freely.

(b) Let H be the maximum height reached by the rocket after the engine stops.

Use III equation of motion

v^2 = u^2 + 2aH

here, v = 0, u = 63.95 m/s, a = - 9.8 m/s^2

0 = 63.95^2 - 2 x 9.8 x H

H = 208.65 m

The maximum height attained by the rocket from the ground is h + H = 140 + 208.65 = 348.65 m

(c) Let t' be the time in which rocket reaches to maximum height after engine is stopped.

Use I equation of motion

v = u + a t'

0 = 63.95 - 9.8 x t'

t' = 6.53 s

The time taken by the rocket to maximum height after lift off is t + t' = 2.32 + 6.53 = 8.85 s.

(d) let t'' be the time taken by the rocket to fall freely

Use II equation of motion

H' = ut'' + 1/2 gt''^2

Here, H' = 348.65 m, u = 0

348.65 = 0 + 0.5 x 9.8 x t''^2

t''^ = 8.44 s

The total time taken by the rocket in air is t + t' + t'' = 2.32 + 6.53 + 8.44 = 17.3 second.