A ball encounters no air resistance when thrown into the air with 100 J of kinetic energy, which is transformed to gravitational potential energy at the top of its trajectory. What is its kinetic energy when it returns to its original height

The kinetic energy when it returns to its original height is 100 J

Explanation:

The ball is thrown up with a Kinetic Energy K. E. = 0.5*m*v² = 100 J

Therefore the final height is given by

u² = v² - 2·g·s

Where:

u = final velocity = 0

v = initial velocity

s = final height

Therefore v² = 2·g·s = 19.62·s

P. E = Potential Energy = m·g·s

Since v² = 2·g·s

Substituting the value of v² in the kinetic energy formula, we obtain

K. E. = 0.5*m*2·g·s = m·g·s = P. E. = 100 J

When the ball returns to the original height, we have

v² = u² + 2·g·s

Since u = 0 = initial velocity in this case we have

v² = 2·g·s and the Kinetic energy = 0.5·m·v²

Since m and s are the same then 0.5·m·v² = 100 J.