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