Question

A rocket is fired vertically upwards with an initial velocity of 92 m/s at the ground level. Its engines then fire, and it is accelerated at 4 m/s until it reaches an altitude of 1200 m. At that point, the engines fail, and the rocket goes into free-fall. Disregard air resistance.

(a) How long was the rocket above the ground?
(b) What is the maximum altitude?
(c) What is the velocity just before it collides with the ground?

Answer 1

The rocket is above the ground for 23 seconds and reaches a maximum altitude of 1200 meters. It does not collide with the ground but continues moving upwards until it reaches the maximum altitude. The initial acceleration of the rocket is 4 m/s.

Step-by-step explanation:

Given:

Initial velocity (u) = 92 m/s

Acceleration (a) = 4 m/s2 (upwards)

Final velocity (v) = 0 m/s (at maximum altitude)

Displacement (s) = 1200 m

(a) To find the time taken to reach the maximum altitude, we can use the second equation of motion:

v2 = u2 + 2as

0 = (92)2 + 2(4)(s)

s = -423 m

The negative value of displacement indicates that the rocket is moving downwards. So, we need to find the time it takes to move 423 meters downwards.

Using the third equation of motion:

v = u + at

0 = 92 + (-4)t

t = 23 seconds

The time taken to reach the maximum altitude is 23 seconds.

(b) To find the maximum altitude, we can use the first equation of motion:

v = u + at

0 = 92 + (-4)t

t = 23 seconds
Using the second equation of motion:

s = ut + (1/2)at2

s = (92)(23) + (1/2)(-4)(23)2

s = 1200 m

The maximum altitude reached by the rocket is 1200 meters.

(c) To find the velocity just before it collides with the ground, we can use the second equation of motion:

v2 = u2 + 2as

v = sqrt(922 + 2(4)(-1200))

v = sqrt(8464 - 9600)

v = sqrt(-1136)

Since the velocity cannot be negative, the rocket will not collide with the ground, but will continue moving upwards until it reaches its maximum altitude.