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amm1812
4 years ago
10

A 0.54-kg ball is thrown vertically upward with an initial speed of 15.5 m/s. What is the potential energy of the ball when it h

as travelled one-quarter of the distance to its maximum height
Physics
1 answer:
Andrew [12]4 years ago
4 0

Answer:

16.2J

Explanation:

At the maximum height v = 0

V ^2= u^2 - 2gh (- because since the ball is going upward)

h = u^2 - v^2 / 2g

15.5^2 /20

h = 12.0125

h = 12m

At one quarter of the distance 1/4 × 12 = 3m

Potential energy = mgh

= 0.54 × 10× 3

= 16.2J

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25 = 0*5  +  (1/2)* a * 5^2.

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3) Mass of the box
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