JEE MAIN - Physics (2022 - 27th June Evening Shift - No. 3)
When a ball is dropped into a lake from a height 4.9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v. It reaches the bottom of the lake 4.0 s after it is dropped. The approximate depth of the lake is :
19.6 m
29.4 m
39.2 m
73.5 m
Explanation
$${t_1} = \sqrt {{{2h} \over g}} $$
$$ = \sqrt {{{2 \times 4.9} \over {9.8}}} = 1\,s$$
$$\Delta t = 4 - 1 = 3\,s$$,
$$v = \sqrt {2gh} = \sqrt {2 \times 9.8 \times 4.9} = 9.8$$ m/s
$$\therefore$$ depth $$ = 9.8 \times 3 = 29.4$$ m
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