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.
