A ball is dropped from a height h above the g | vedclass

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Question

(A) Let the ball be dropped from a height $h$ above the ground.When the ball has fallen through a distance $d = (h - y)$,where $y$ is the height above

Vedclass · Accepted Answer

$\sqrt{2g(h-y)}$

Vedclass · Answer

(A) Let the ball be dropped from a height $h$ above the ground.When the ball has fallen through a distance $d = (h - y)$,where $y$ is the height above the ground,its velocity $v$ can be calculated using the third equation of motion: $v^2 = u^2 + 2as$.Here,initial velocity $u = 0$,acceleration $a = g$,and the distance covered $s = h - y$.Substituting these values,we get: $v^2 = 0^2 + 2g(h - y)$.Therefore,the velocity $v$ as a function of height $y$ is $v = \sqrt{2g(h - y)}$.

$A$ ball is dropped from a height $h$ above the ground. Neglecting air resistance,its velocity $(v)$ varies with its height $(y)$ above the ground as:

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