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 re...
Question 1 Report
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.
Answer Details
A body of mass \(m = 0.6\,\text{kg}\) is thrown vertically upward from the ground with initial speed \(u = 20\,\text{ms}^{-1}\).
(i) Potential energy at maximum height
By conservation of energy, at the highest point all the initial kinetic energy has become potential energy (the body is momentarily at rest):
\[ \text{P.E.}_{max} = \frac{1}{2}mu^{2} = \frac{1}{2}(0.6)(20)^{2} = \frac{1}{2}(0.6)(400) = 120\,\text{J} \](ii) Kinetic energy just before it hits the ground
Neglecting air resistance, the body returns to the ground with the same speed it left, so its kinetic energy just before impact equals its initial kinetic energy (equal also to the P.E. at the top):
\[ \text{K.E.} = \frac{1}{2}mu^{2} = 120\,\text{J} \]Download The App On Google Playstore
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