If a body mass 5kg is thrown vertically upwards with velocity u,at what height will the potential energy equal to the kinetic energy?
From V\(^2\) = U\(^2\) + 2as
O = U\(^2\) - 2gH, since at maximum height H, the final velocity, v = O.
when K.E = P.E
∴ H = \(\frac{(U)^2}{2g}\)
Halfway is the \(\frac{\text{height}}{2}\)
PE = KE at \(\frac{(1)}{2}\) H = \(\frac{(U)^2}{2g}\) x (\(\frac{1}{2}\)) = \(\frac{U^2}{4g}\)
There is an explanation video available below.
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