Question

A body dropped from top of a tower fall through $40m$ during the last two seconds of its fall. The height of tower is $(g=10m/s^2)$

• A. 60 m
• B. 45 m
• C. 80 m
• D. 50 m

Answer 1

Answer: (B)

Let $h$ be height of the tower and $t$ is the time taken by the body to reach the ground.
Here, $u=0,a=g$
$\therefore h=ut+\frac{1}{2}g{t}^2$
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 $40=\frac{1}{2}g{t}^2-\frac{1}{2}g\left(t^2+4-4t\right)$
or $40=(2t-2)g$ or $t=3s$
From eqn (i), we get $h=\frac{1}{2}\times 10\times (3{)}^2$
or $h=45m$