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 balls collide exactly midway. What is the speed in m/s with which the second ball is thrown?

• A. $31.3$
• B. $27.8$
• C. $22.4$
• D. $19.6$

Answer 1

Answer: (A)

$31.3$

Time $t$ for dropped ball to reach $50m$ distance from top can be obtained from,

$S=\frac{1}{2}g{t}^2$, where $S$ is the distance $50m$ and $g$ is acceleration due to gravity

Hence, we have $t=(\frac{2S}{g}{)}^{\frac{1}{2}}=(\frac{100}{9.8}{)}^{\frac{1}{2}}\approx 3.2s$

Now in this $3.2s$ , the ball thrown from bottom should reach $50m$.

We have the relation, $S=u\times t-\left(\frac{1}{2}\right)\times g\times t^2$ ..............................(1)

where $S$ is the distance $50m$ and $u$ is the initial speed of ball thrown from bottom

Substituting the values in eqn.(1), we get $u=31.3m/s$