A stone falls from rest. The total distance covered by it in the last second of its motion is equal to the distance covered in the first three seconds. What is the height from which the stone was dropped? Take g = 10 $m / s^{2}$ :

45 m

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100 m

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125 m

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200 m

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Solution

The correct option is C 125 m
Distance covered in 3 sec $= \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times 3^{2} = 45 m$
At the end of 'n' second its velocity $u = g n$
At the end of 'n+1' second its velocity $v = g (n + 1)$

Distance traveled in the (n+1)th second $= \frac{(2 n + 1) g}{2} = \frac{(20 n + 10)}{2} = 10 n + 5$

$10 n + 5 = 45 \Rightarrow (n + 1) = 5$
Distance covered in 5 second $= \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times 5^{2} = 125 m$

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A stone falls from rest. The total distance covered by it in the last second of its motion is equal to the distance covered in the first three seconds. What is the height from which the stone was dropped? Take g = 10 m/s2:

The correct option is C 125 mDistance covered in 3 sec=12gt2=12×10×32=45mAt the end of 'n' second its velocity u=gnAt the end of 'n+1' second its velocity v=g(n+1)Distance traveled in the (n+1)th second=(2n+1)g2=(20n+10)2=10n+510n+5=45⇒(n+1)=5Distance covered in 5 second =12gt2=12×10×52=125m

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