Question

From the top of a tower of height $50m$, a ball is thrown vertically upwards with a certain velocity. It hits the ground $10s$ after it is thrown up. How much time does it take to cover a distance $AB$ where $A$ and $B$ are two points $20m$ and $40m$ below the edge of the tower? ( $g=10m{s}^{-2}$ )

• A. 2.0 s
• B. 1.0 s
• C. 0.5 s
• D. 0.4 s

Answer 1

Answer: (D)

Let the body be projected upwards with velocity $u$ from top of tower. Taking vertical downward motion of body form top of tower to ground, we have
$u=-u,a=g=10m{s}^{-2},s=50m,t=10s$
As $s=ut+\frac{1}{2}a{t}^2$ ,
So, $50=-u\times 10+\frac{1}{2}\times 10\times 10^2$
On solving $u=45m{s}^{-1}$
If $t_1$ and $t_2$ are the timings taken by the ball to reach points $A$ and $B$ respectively, then
$20=45t_1+\frac{1}{2}\times 10\times t_1^2$
and $40=-45t_2+\frac{1}{2}\times 10\times t_2^2$
On solving, we get $t_1=9.4s$ and $t_2=9.8s$
Time taken to cover the distance $AB$
$=\left(t_2-t_1\right)=9.8-9.4=0.4s$.