A body falls freely from top of a tower. It covers 36% of total height in last second before strikin

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2.Motion in Straight Line

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A body falls freely from the top of a tower. It covers $36\%$ of the total height in the last second before striking the ground level. The height of the tower is........$m$

A

$50$

B

$75$

C

$100$

D

$125$

Solution

(d) Let height of tower is $h$ and body takes t time to reach to ground when it fall freely.

$\therefore $ $h = \frac{1}{2}g\,{t^2}$…$(i)$

In last second i.e. ${t^{th}}$ sec body travels $= 0.36 h$

It means in rest of the time i.e. in $(t – 1)$ sec it travels

$ = h – 0.36\;h = 0.64\;h$

Now applying equation of motion for $(t -1) \,sec$

$0.64\,h = \frac{1}{2}g\,{(t – 1)^2}$…$(ii)$

From $(i)$ and $(ii)$ we get, $t = 5\sec $ and $h = 125m$

Standard 11

Physics

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