JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 10)
A child stands on the edge of the cliff $$10 \mathrm{~m}$$ above the ground and throws a stone horizontally with an initial speed of $$5 \mathrm{~ms}^{-1}$$. Neglecting the air resistance, the speed with which the stone hits the ground will be $$\mathrm{ms}^{-1}$$ (given, $$g=10 \mathrm{~ms}^{-2}$$ ).
20
25
30
15
Explanation
_1st_February_Morning_Shift_en_10_1.png)
$$ \begin{aligned} & \mathrm{v}_{\mathrm{y}}=\sqrt{2 g h}=\sqrt{200} \\\\ & v_{n e t}=\sqrt{25+200}=15 \mathrm{~m} / \mathrm{s} \end{aligned} $$
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