An object is moving with a velocity of 5m-1. At what height must a similar body be situated to have a potential energy equal in value with kinetic energy of the moving body?
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Correct Answer: Option C
Explanation:
P.E = K.E
Therefore, h = \(\frac{v^2}{2g} = \frac{5^2}{2 \times 10}\)
= 1.25
Approx 1.3 (C)
P.E = K.E
Therefore, h = \(\frac{v^2}{2g} = \frac{5^2}{2 \times 10}\)
= 1.25
Approx 1.3 (C)