WAEC - Physics (2013 - No. 9)
A body of mass 20g projected vertically upwards in vacuum returns to the point of projection after 1.2s. [g = 10ms-2]. Determine the potential energy of the body at the maximum height of its motion
0.36J
0.72J
360.00J
720.00J
Explanation
P.E at max = K.E on projection
= \(\frac{1}{2}\)mv2 = \(\frac{1}{2} \times 0.02 \times 6^2 = 0.36J\)
= \(\frac{1}{2}\)mv2 = \(\frac{1}{2} \times 0.02 \times 6^2 = 0.36J\)
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