Arocket is launched from ground level with the constant upward acceleration of 20 m/sec. after 10 seconds the fuel is exhausted and the rocket coasts up against gravity until it reaches the highest point, then falls back to earth. how high does it go and what is the velocity when it returns to ground level?

Explanation:

Given:

Rocket launched with

acceleration, a = +20 m/s^2 (upwards)

for

time, t = 10 seconds

coasts against gravity and eventually returns to ground.

Need

1. Maximimum height, H, reached

2. velocity on landing.

Solution:

We need to use the following kinematics equations:

v = u + at (1)

v^2-u^2 = 2aS (2)

where

u = initial velocity (m/s, of a particular stage)

v = final velocity (m/s)

a = acceleration (m/s^2)

S = distance travelled (m)

The rocket goes through three stages.

A. with acceleration upwards

B. no more propulsion, upwards movement (in deceleration)

C. no more propulsion, free-fall from highest point.

Stage A: with acceleration until fuel exhausted

u = 0 m/s

a = 20 m/s^2

t = 20 s

to find final velocity (at end of stage 1, apply equation (1):

v =  u+at = 0 + 20*20 = 400 m/s

to find height reached, apply equation (2)

v^2-u^2 =  2aS  (2)  =>

S = (v^2-u^2)/2a (2a)

= (400^2-0^2) / (2*20)

= 4000 m  (above ground)

Stage B: coasting upwards

u = 400 m/s  (from stage A)

v = 0 (at highest point)

a = -9.81 (acceleration due to gravity, downwards, so negative)

to find distance travelled during stage B, apply equation (2a)

S = (v^2-u^2)/2a

= (0^2 - 400^2) / (2*-9.81)

= 8154.94 m

Stage C :  free-fall from highest point to ground

At highest point, Total upward distance travelled (stages A & B)

= 4000 + 8154.94 = 12154.94 m   (Q1)

u = 0

a = -9.81 m/s^2

S = (4000+8154.94) = 12154.94

To find velocity on impact at ground level, use equation (2)

v^2 - u^2 = 2aS  (2)    =>

v = sqrt(2aS+u^2)  (2b)

= sqrt(2*20*12154.84+0^2)

= sqrt(486197.76)

= 697.3 m/s   (Q2)

i think the answer is b