Question

A bowling ball is traveling at 3 m/s. It starts to roll up a ramp. How high above the ground will the ball be when it stops rolling? Neglect friction and assume the ramp is plenty long enough to do this. (For this problem your initial kinetic energy can be set equal to your final potential energy. 1/2mv2 = mgh).

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

h= 0.45 m

Explanation:

PE= 1/2 mv^2             KE= mgh

v= 3m/s

vf= 0 m/s

h=?

PE= 1/2(1kg)(3m/s)^2

PE= 4.5 J

4.5 J/ 1kg(9.8 m/s^2)

h=0.45 m