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 has travelled one-quarter of the distance to its maximum height

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