Question

A ball $A$ is dropped from a $44.1\text{m}$ high cliff. Two seconds later, another ball $B$ is thrown downward from the same place with some initial speed. The two balls reach the ground together. Find the speed with which the ball $B$ was thrown.
$[\text{Take}g=9.8{\text{m/s}}^2]$

• A. $39.2\text{m/s}$
• B. $44.1\text{m/s}$
• C. $24.6\text{m/s}$
• D. $12.6\text{m/s}$

Answer 1

The correct option is A $39.2\text{m/s}$

Time taken by the ball $A$ to reach the ground is given by
$44.1=\frac{1}{2}\times 9.8\times t^2$
or $t^2=\frac{2\times 44.1}{9.8}=9$
or $t=3\text{s}$

Suppose, the ball $B$ was thrown with initial speed $u$ downward. It was thrown $2\text{s}$ after $A$ was dropped, and it reached the ground together with $A$.

So, the total time for which it has fallen is $3-2=1\text{s}$.

In this time, the ball $B$ has covered a distance of $44.1\text{m}$.

We have,
$s=ut+\frac{1}{2}g{t}^2$
$\Rightarrow 44.1=u\times 1+\frac{1}{2}\times 9.8\times 1^2$
or $u=39.2\text{m/s}$