Question

A ball is projected upwards from the foot of a tower. The ball crosses the top of the tower twice after an interval of $6s$ and the ball reaches the ground after $12s$. The height of the tower is ($g=10m/s^2$):

• A. 120m
• B. 135m
• C. 175m
• D. 80m

Answer 1

The correct option is B 135m

Let height of tower be $h$.

As given,

$t_{BCD}=6sec$

and $t_{ABCDE}=12sec$

$\therefore t_{AB}=3sec=t_{DC}$

and , $\therefore t_{BC}=3sec=t_{CD}$

So, from relation $v=u+at$

From $A$ to $C,u=U$

$v=0,a=-g,t=6sec$

$O=u-gt,t=\frac{u}{g},$ or $u=10\times 6m/s$

$u=60m/s$

Now, from relation $s=ut+\frac{1}{2}a{t}^2$

from, $A$ to $B,s=h,u=60m/s,t=3sec,a=-g$

$h=60\times 3-\frac{1}{2}\times 10\times 9$

$h=135m$