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

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Question

(B) The time taken by any body of mass $m$ to reach the ground when dropped from a height $h$ is given by the equation of motion $h = ut + \frac{1}{2}

Vedclass · Accepted Answer

$t_1 = t_2$

Vedclass · Answer

(B) The time taken by any body of mass $m$ to reach the ground when dropped from a height $h$ is given by the equation of motion $h = ut + \frac{1}{2}gt^2$. Since the initial velocity $u = 0$,we have $h = \frac{1}{2}gt^2$.Solving for $t$,we get $t = \sqrt{\frac{2h}{g}}$.Since the time $t$ depends only on the height $h$ and the acceleration due to gravity $g$,it is independent of the mass of the object.Therefore,for two balls dropped from the same height,the time taken 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 equal height. If the time taken by the balls are $t_1$ and $t_2$ respectively,then:

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