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Q. 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

Haryana PMTHaryana PMT 2010Motion in a Straight Line

Solution:

Let height of tower is h and body takes time t to reach the ground when it falls freely.
$\therefore h=\frac{1}{2} g t^{2} \ldots$(i)
In last second ie, the sec body travels $=0.36 h$
It means in rest of the time ie, 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} \ldots $ (ii)
From Eqs. (i) and (ii), we get
$t=5 \,s$ and $h=125\, m$