A. h =(u)²/g
B. h = (u)²/4g
C. h = (2u)²/g
D. h = (u)²/2g
Correct Answer:
Option B – h = (u)²/4g
Explanation
From V² = U² + 2as
O = U² – 2gH, since at maximum height H, the final velocity, v = O.
∴ H = (U)²/2g
PE = KE at (1)/2 H = (U)²/2g x (1)/2 = U²/4g
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![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.](https://erudites.ng/wp-content/plugins/contextual-related-posts/default.png "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.")