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.
Explanation
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.