Question

A ball is dropped from the top of the building $100m$ high. Simultaneously, another ball is thrown upwards from the bottom of the building with such a velocity that the balls collide exactly mid-way. What is the speed in $m{s}^{-1}$ with which the second ball is thrown? (Take $g=10m{s}^{-2}$)

• A. 31.6
• B. 27.8
• C. 22.4
• D. 19.6

Answer 1

Answer: (A)

For first ball,
Time taken to reach mid-way
$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 50}{10}}=\sqrt{10}s$ $[\because h=\frac{100}{2}m]$
For second ball, using $v=u+at$
$v=0,a=-g,u=?$
Let $t$' be time taken by the ball to reach midway.
$\therefore 0=u-g{t}^{\prime }or{t}^{\prime }=\frac{u}{g}=\frac{u}{10}$
As both the balls collide at same time so
$(t-0)=(t^{\prime }-0)ort=t^{\prime }$
$\sqrt{10}=\frac{u}{10};u=10\sqrt{10}=31.6m{s}^{-1}$