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 $a}_{y}=12.80\frac{m}{{s}^{3}$ )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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Daniel Williams

Answered 2021-11-19
Author has **14** answers

a) Since the acceleration is constant constant, integrate the first kinematics equation:

with,

b) Substitute y=325 m in the equation:

then substitute this time in the equation:

Result: a) y=467 m

b)

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t=N

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