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When a ball is dropped anto a lake from a height 4.9 \mathrm{~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 \mathrm{~s}after it is dropped. The approximate depth of the lake is :

Option: 1

19.6 \mathrm{~m}


Option: 2

29.4 \mathrm{~m}


Option: 3

39.2 \mathrm{m}


Option: 4

73.5 \mathrm{m}


Answers (1)

best_answer


\mathrm{v= \sqrt{2gh}= \sqrt{2\times 9.8 \times 4.9}= 9.8\, m/s}
\mathrm{\text{time for A to B}= \sqrt{\frac{2h}{g}}= \sqrt{\frac{2\times4.9}{9.8}}= 1\, sec}
distance travelled by ball in 3 sec in lake \mathrm{= 9.8\times 3}
\mathrm{\therefore }  depth of lake \mathrm{= 29.4\, m}

Correct answer is (2)

Posted by

Nehul

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