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.
(i) potential energy at the maximum height reached.
(ii) kinetic energy just before it hits the ground.
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Correct Answer: Option n
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.
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.