Question

A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 joules (with the potential energy equal to zero at ground level) and is moving upward with a kinetic energy of 50 joules. Air friction is negligible. The maximum height reached by the ball is most nearly

Answer 1

Answer: 20m

Step-by-step explanation:

We will solve this question by applying the law of conservation of energy which states that the sum of potential energy and kinetic energy is always the same.

The PE is 0 at surface and maximum at top while the KE is maximum at surface and 0 at top.

From the question,

PE = mgh = 50 J -(1)

mg* 10 = 50

mg = 50/10

mg = 5

The total energy at that point = PE + KE = 50 + 50 = 100 J

Therefore, at topmost point, the PE will be 100 J

mgH = 100J , H is the needed height

Using the value of mg obtained above, we have

H= 100/5

H = 20 m