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how much potential energy does the ball have when it reaches the top of its ascent​

Sagot :

PE = 450 m

Further explanation

Given

The initial problem might be like this

A 1.0-kg ball is thrown into the air with an initial velocity of 30 m/s.

Required

Potential energy

Solution

h max = vo²sin²θ/2g

h max = 30²sin²90/2.10

h max = 45 m

PE = m.g.h

PE = 1 x 10 x 45

PE = 450 m

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