A body of mass 0.6kg is thrown vertically upward from the ground with a speed of 20ms\(^{-2}\). Calculate its;
(i) potential energy at the maximum height reached.
(ii) kinetic energy just before it hits the ground.
Kinetic energy at point of projection = \(\frac{1}{2} mv^2 \times 0.6 \times 20^2\) = 120J
or
V\(^2\) = U\(^2\) - 2gs
S = \(\frac{v^2 - U^2}{-2g}\) = \(\frac{0^2 - 2062}{-2 \times 10}\)
= 20cm
P.E. = mgs or mgh 0.6 x 10 x 20 = 120J
ii) Kinetic energy in reaching ground = Potential energy at highest point = 120J.
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