A rocket starts from rest and moves upward from the surface of the earth. For th

Question

A rocket starts from rest and moves upward from the surface of the earth. For the first 10s of its motion, the vertical acceleration of the rocket is given by ay = (2.80 m/s^3) t, where the +y - direction is upward.

1.) What is the height of the rocket above the surface of the earth at t= 10s?

I know this answer is h = 467 m

But, I can't figure the second question...

2.) What is the speed of the rocket when it is 295 m above the surface of the earth?

Express your answer with the appropriate units.

Vy = ??

I can't figure out how to do the second question. Please show step by step with correct answers so I can try to figure it out.

THANK YOU!!!

Explanation / Answer

a = d2h/dt2 = 2.8t

v = dh/dt = 1.4t^2 +c1

at t = 0 v = 0 so c1 = 0

h = 0.4667t^3 = 0.4667*10^3 = 466.7 m

0.4667t^3 = 295

t = 8.589 sec

v = 1.4t^2 = 103.28 m/s

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