A ball of mass m_1 and another ball of mass m | vedclass

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(B) The motion of an object under gravity is governed by the equation of motion $h = ut + \frac{1}{2}gt^2$. Since the balls are dropped from rest,the

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$t_1 = t_2$

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(B) The motion of an object under gravity is governed by the equation of motion $h = ut + \frac{1}{2}gt^2$. Since the balls are dropped from rest,the initial velocity $u = 0$. Therefore,the equation becomes $h = \frac{1}{2}gt^2$. Solving for time $t$,we get $t = \sqrt{\frac{2h}{g}}$. This expression shows that the time of fall $t$ depends only on the height $h$ and the acceleration due to gravity $g$. It is independent of the mass of the object. Since both balls are dropped from the same height $h$,the time taken by both balls will be equal,i.e.,$t_1 = t_2$.

$A$ ball of mass $m_1$ and another ball of mass $m_2$ are dropped from the same height. If the time taken by the balls to reach the ground is $t_1$ and $t_2$ respectively,then:

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