Question

A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 s another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v?(Take $g=10m{s}^{-2}$ )

• A. $74m{s}^{-1}$
• B. $64m{s}^{-1}$
• C. $84m{s}^{-1}$
• D. $94m{s}^{-1}$

Answer 1

Answer: (A)

For first ball, $u=0$ $\therefore$ $s_1=\frac{1}{2}g{t}_1^2=\frac{1}{2}\times g{(18)}^2$ For second ball, initial velocity $=v$ $\therefore$ $s_2=v{t}_2+\frac{1}{2}\times g{t}^2$ $t_2=18-6=12s$ $\Rightarrow$ $s_2=v\times 12+\frac{1}{2}g{(12)}^2$ Here, $s_1=s_2$ $\frac{1}{2}g{(18)}^2=12v+\frac{1}{2}g{(12)}^2$ $\Rightarrow$ $v=74m{s}^{-1}$