A ball is dropped from a high rise platform at $\mathrm{t}=0$ starting from rest. After $6 \mathrm{~s}$…

A ball is dropped from a high rise platform at $\mathrm{t}=0$ starting from rest. After $6 \mathrm{~s}$ another ball is thrown downwards from the same platform with a speed $\mathrm{v}$. The two balls meet at $\mathrm{t}=18 \mathrm{~s}$. What is the value of $\mathrm{v}$ ? (take $g=10 \mathrm{~ms}^{-2}$ )
  1. $74 \mathrm{~ms}^{-2}$
  2. $55 \mathrm{~ms}^{-1}$
  3. $40 \mathrm{~ms}^{-1}$
  4. $60 \mathrm{~ms}^{-1}$

Solution

. For first ball, $\mathrm{u}=0$ $\therefore \quad \mathrm{s}_1=\frac{1}{2} \mathrm{gt}_1^2=\frac{1}{2} \times \mathrm{g}(18)^2$ For second ball, initial velocity $=\mathrm{v}$ $\begin{aligned} & \therefore \mathrm{s}_2=\mathrm{vt}_2+\frac{1}{2} \mathrm{gt}^2 \\ & \Rightarrow \quad \mathrm{t}_2=18-6=12 \mathrm{~s} \\ & \Rightarrow \quad \mathrm{s}_2=\mathrm{v} \times 12+\frac{1}{2} \mathrm{~g}(12)^2 \end{aligned}$ Here, $\quad \mathrm{s}_1=\mathrm{s}_2$ $\begin{aligned} \frac{1}{2} g(18)^2 & =12 \mathrm{v}+\frac{1}{2} g(12)^2 \\ \Rightarrow \quad v & =74 \mathrm{~ms}^{-1} \end{aligned}$

Asked in: NEET 2010 (Screening)

Practice more Motion In One Dimension questions on Aicharya