Question

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

• 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
$u=-u,a=g=10\,ms^{- 2},s=50\,m,t=10\, s$
As $s=ut+\frac{1}{2}at^{2}$ ,
So, $50=-u\times 10+\frac{1}{2}\times 10\times 10^{2}$
On solving $u=45\,ms^{- 1}$
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 $t_{1}=9.4 \, s$ and $t_{2}=9.8\, s$
Time taken to cover the distance $AB$
$=\left(t_{2} - t_{1}\right)=9.8-9.4=0.4 \, s$.