Question

A body dropped from top of a tower fall through $60m$ during the last two seconds of its fall. The height of tower is : $(g=10m/s^2)$

• A. $95m$
• B. $60m$
• C. $80m$
• D. $90m$

Answer 1

Answer: (C)

Let body reaches the ground in $t$ seconds.
$\therefore$ Velocity of body after $(t-2)$ seconds from equation of motion

$v=u+g{t}^{\prime }$
where $v$ is final velocity, $u$ the initial velocity
$(=0)$ and $t^{\prime }=t-2$
$\therefore v=g(t-2)$
Distance covered in last two seconds
$h^{\prime }=g(t-2)\times 2+\frac{1}{2}g(2{)}^2$
$60=20(t-2)+20$
On solving $t=4s$
Hence, height of tower is given by
$h=ut+\frac{1}{2}g{t}^2$
Since, $u=0$
$\therefore h=\frac{1}{2}g{t}^2$
$=\frac{1}{2}\times 10\times (4{)}^2=80m$