Question

A rocket moves upward starting from rest with an acceleration of 20 m/s^2 for 5 s. It runs out of fuel at the end of this 5 s and continues to move upward, acted upon only by gravity. How high does it rise? Find the velocity of the rocket just before it hits the ground and how long it is in the air?

Answer 1

a)

Hmax = distance trvalled during constant acceleration and distance trvelled after engine is off

Hmax = 0.5 x20x5x5 + (20x20x5x5/2g)

Hmax = 760.2 m

b)

at maximum height velocity = 0

therefore we have

V^2 -u^2 = 2x g x Hmax

V = 122.065m/s

Answer 2

SOLUTION :

a.

Velocity, v(5)  after 5 sec. = a t = 20*5 = 100 m/s

Average velocity during this interval = (0 + 100)/2 = 50 m/s

As the acceleration during this interval is constant, average velocity can be considered as constant velocity during the period.

So, height H(5) attained in 5 sec. 

= average velocity * time

= 50*5

= 250 m 

After 5 sec. engine is turned off and gravity comes into play.

So, 

Initial velocity = v(5) = 100m/s.

Gravitational acceleration (opposite to the up movement) = - g = - 9.8 m/s^2

At he maximum height, Hm , velocity is 0 m/s.

Let it takes time t sec. to reach the highest point.

So,

v(t) = v(5) - g t 

=> 0 = 100 - 9.8 t 

=> t = 100/9.8 = 10.2041 s

Average velocity in the in the duration of t sec = (100+0)/2 = 50 m/s

Height attained in t sec, H(t) 

= average velocity * t 

= 50 t 

= 50 * 10.2041

= 510.20 m

So,

Maximum height, Hm

= H(5) + H(t) 

= 250 + 510.20

= 760.20 m (ANSWER).

b.

From maximum height it starts falling. 

Initial velocity = 0 m/s

Gravitational acceleration = g = 9.8 m/s^2

Distance of fall = 760.20 m

Let it take t’ sec to fall to ground and v(t’) be the velocity I t’ sec.

So,

v(t’) = 0 + g t’ = 0 + 9.8 t’ = 9.8 t’ 

Average velocity = (0 + v(t’))/2 = (0 + 9.8 t’)/2 = 4.9 t’

Distance travelled in t’ sec = Avg. velocity * t’ 

=> 760.20 = 4.9 t’ * t’ = 4.9 t’^2

=> t’ = sqrt(760.20/4.9) = 12.4556 s

So, 

Velocity while hitting the ground, v(t’)

= 9.8 * 12.4556

= 122.06 m/s (ANSWER)