Question

A ball is dropped from the top of the building $100 \,m$ 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 $g = 10\, m \,s^{-2}$)

• 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 \times50}{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.
$\therefore \:\:\:\: 0 = u - gt' \:\: or \:\:\: t' = \frac{u}{g} = \frac{u}{10}$
As both the balls collide at same time so
$(t - 0) = (t' - 0) \, or \, t = t'$
$\sqrt{10} = \frac{u}{10} ; u = 10 \sqrt{10} = 31.6 \, m \, s^{-1}$