A ball of mass 2;kg and another of mass 4;kg | vedclass

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Question

(C) When objects are dropped from the same height under gravity,their velocity $v$ after falling a distance $h$ is given by the equation of motion $v^

Vedclass · Accepted Answer

$1 : 2$

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(C) When objects are dropped from the same height under gravity,their velocity $v$ after falling a distance $h$ is given by the equation of motion $v^2 = u^2 + 2gh$. Since both balls start from rest $(u = 0)$,the velocity after falling $30\;ft$ is $v = \sqrt{2gh}$.Since $g$ and $h$ are the same for both balls,their velocities are equal $(v_1 = v_2)$.The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2}mv^2$.Therefore,the ratio of their kinetic energies is $\frac{K.E._1}{K.E._2} = \frac{\frac{1}{2}m_1v_1^2}{\frac{1}{2}m_2v_2^2} = \frac{m_1}{m_2}$.Given $m_1 = 2\;kg$ and $m_2 = 4\;kg$,the ratio is $\frac{2}{4} = \frac{1}{2}$.

List-$I$	List-$II$
$(A)$ $P(E_2)$	$(i)$ $1/4$
$(B)$ $P(E_1 \cup E_2)$	$(ii)$ $5/8$
$(C)$ $P(\bar{E}_1 / \bar{E}_2)$	$(iii)$ $1/8$
$(D)$ $P(E_1 / \bar{E}_2)$	$(iv)$ $1/2$
	$(v)$ $3/8$
	$(vi)$ $3/4$

$A$ ball of mass $2\;kg$ and another of mass $4\;kg$ are dropped together from a $60\;ft$ tall building. After a fall of $30\;ft$ each towards the Earth,their respective kinetic energies will be in the ratio of?

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