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

sklicatias 2021-11-18 Answered
A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by ay=12.80ms3 )t, where the + y-direction is upward. (a) What is the height of the rocket above the surface of the earth at t = 10.0 s? (b) What is the speed of the rocket when it is 325 m above the surface of the earth?
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Expert Answer

Daniel Williams
Answered 2021-11-19 Author has 14 answers

a) Since the acceleration is constant constant, integrate the first kinematics equation:
vy=0taydt=0102.8tdt=1.4t2+v0
y=0tvdt
with,v0=0 and y0=0,
y=0101.4t2dt=0.467t3010=467 m
b) Substitute y=325 m in the equation:
y=vdt=1.4t2dt=0.467t3
325=0.476t3
t=8.86 s
then substitute this time in the equation:
vy=0taydt=08.862.8dt=1.4t208.86=110 m
Result: a) y=467 m
b) vy=110 m

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