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(A) For a ball thrown vertically upwards:Initial velocity $= u 	ext{ m/s}$.Final velocity at maximum height $= v = 0 	ext{ m/s}$.Using the equation

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$u$

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(A) For a ball thrown vertically upwards:Initial velocity $= u 	ext{ m/s}$.Final velocity at maximum height $= v = 0 	ext{ m/s}$.Using the equation of motion $v = u + at$,where $a = -g$ (acceleration due to gravity acting downwards):$0 = u - gt \implies t = u/g$ (Time taken to reach maximum height).For the return journey,the ball starts from rest at the maximum height $(u_{return} = 0)$ and falls downwards under gravity $(a = +g)$:Using the equation $v^2 = u^2 + 2as$:$v^2 = 0^2 + 2gH$,where $H$ is the maximum height.From the upward journey,$H = u^2 / (2g)$.Substituting $H$ in the return equation: $v^2 = 2g(u^2 / 2g) = u^2$.Therefore,$v = u$.The ball returns to the point of projection with the same velocity $u$ with which it was thrown.

$A$ ball is thrown vertically upwards with a velocity $u$. Calculate the velocity with which it falls to the earth again.

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