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
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Let height of tower is h and body takes t time to reach to ground when it fall freely.
\(h = \frac{1}{2} g t^{2}\) …(i)
In last second i.e. \(\{ L^{\infty} \}\) sec body travels = 0.36 h
It means in rest of the time i.e. in \{t - 1\} sec it travels
\((h - 0.36 h \quad 0.64 h)\)
Now applying equation of motion for ( t – 1) sec
\(0.64h - \frac{1}{2} g (t - 1)^2\) …(ii)
From (i) and (ii) we get, \(\ell = 5 \text{ sec}\) and \(\hbar = 12.5 \pi\)
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