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=(2.80m/s^3)t, where the +y-direction is upward.

Part A
What is the height of the rocket above the surface of the earth at t = 10s ?
Express your answer with the appropriate units.

h = Value Unit

Part B
What is the speed of the rocket when it is 280m above the surface of the earth?
Express your answer with the appropriate units.

vy = Value Unit

Explanation / Answer

a)ay=2.8t m/s^3

s=u*t+0.5*ay*t^2

u=0

h=0.5*2.8*t*t^2=1.4*t^3

height after 10s

h=1.4*10^3=1400m

b)

here rocket is 280m above the earth

so time taken to reach 280m is t=(2h/ay)=(2*280/2.8t)

t^2=2*280/2.8*t

t^3=200

t=5.8s

now velocity at a height 280m

v^2-u^2=2ay*s

v^2-0=2*2.8*t*280

t=5.8s

v^2=2*2.8*5.8*280

v=95.75m/s

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