You have designed a rocket to be used to sample the local atmosphere for polluti

Question

You have designed a rocket to be used to sample the local atmosphere for pollution. It is fired vertically with a constant upward acceleration of 16 m/s2. After 26 s, the engine shuts off and the rocket continues rising (in freefall) for a while. (Neglect any effects due to air resistance.) The rocket eventually stops rising and then falls back to the ground. You want to get a sample of air that is 17 km above the ground.

(a) What is the highest point your rocket reaches?

(b) Determine the total time the rocket is in the air.

(c) Find the speed of the rocket just before it hits the ground.

Explanation / Answer

a)

for initial acceleraon

a = 16 m/sec2

t = 26 sec

initial velcootict u = 0

final velcotity = v ( let )

so.. v = u+at = 0 + 16*26 = 416 m/sec

distance travelled during acceleration s1 = ut + 0.5*a*t^2

so.. s1 = 0 + 0.5*16*26^2 = 5408 m

now... after the engine shuts..

inital velocity u = 416 m/sec

final velocity v = 0

acceletaion = -9.81 m/sec2

so... time taken to reach maximum height = (v - u ) / a = ( 0 - 416 )/(-9.81 ) = 42.40570846 sec

distance moved after engine shuts off till it reaches maximum height = s2 ( let )

so.. v^2 = u^2 + 2*a*s2

so.. 0 = 416^2 - 2*9.81*s2

so.. s2 = 8820.38736 m

so... total height of heighest point = s1 + s2 = 5408 + 8820.38736 = 14228.38736 m

b)

so.. time till the rocket reaches maximum height = t1 = 42.40570846 + 26 = 68.40570846 m

now... for falling back to ground ..

time t = t2

distance = s = 14228.38736 m

intiial velocity u = 0

acceleration a = 9.81 m/sec2

so.. s = ut + 0.5*a*t^2

so.. 14228.38736 = 0 + 0.5 * 9.81 * t2^2

so.. t2 = 53.859 sec

so.. total time taken = t1 + t2 = 53.859 + 68.40570846 = 122.2647145 sec

c) speed of rocket just before hiting = v = u + at = 0+9.81*53.859 = 528.35679 m/sec

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