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Q.
A body dropped from top of a tower fall through $40\, m$ during the last two seconds of its fall. The height of tower is $(g = 10\, m/s^2)$
AIPMTAIPMT 1992Motion in a Straight Line
Solution:
Let $h$ be height of the tower and $t$ is the time taken by the body to reach the ground.
Here, $u=0, a=g$
$\therefore h=u t+\frac{1}{2} g t^{2} $
or $h=0 \times t+\frac{1}{2} g t^{2}$
or $h=\frac{1}{2} g t^{2}\,\,\,\,....$(i)
Distance covered in last two seconds is
$40=\frac{1}{2} g t^{2}-\frac{1}{2} g(t-2)^{2} ($ Here, $u=0$ )
or $ 40=\frac{1}{2} g t^{2}-\frac{1}{2} g\left(t^{2}+4-4 t\right)$
or $40=(2 t-2) g $ or $t=3 s$
From eqn (i), we get $h=\frac{1}{2} \times 10 \times(3)^{2}$
or $h=45 \,m$