JEE MAIN - Physics (2022 - 29th July Evening Shift - No. 12)
A ball is released from a height h. If $$t_{1}$$ and $$t_{2}$$ be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between $$t_{1}$$ and $$t_{2}$$.
$$t_{1}=(\sqrt{2}) t_{2}$$
$$t_{1}=(\sqrt{2}-1) t_{2}$$
$$t_{2}=(\sqrt{2}+1) t_{1}$$
$$t_{2}=(\sqrt{2}-1) t_{1}$$
Explanation
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For first $${h \over 2}$$ distance,
$${h \over 2} = {1 \over 2}~gt_1^2$$
$$ \Rightarrow h = gt_1^2$$
For total distance h,
$$h = {1 \over 2}g{({t_1} + {t_2})^2}$$
$$ \Rightarrow gt_1^2 = {1 \over 2}g{({t_1} + {t_2})^2}$$
$$ \Rightarrow 2t_1^2 = {({t_1} + {t_2})^2}$$
$$ \Rightarrow \sqrt 2 {t_1} = {t_1} + {t_2}$$
$$ \Rightarrow (\sqrt 2 - 1){t_1} = {t_2}$$
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