Q100 / 200 · NEET 2010 (Screening) · Physics

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 $g=10 \mathrm{~ms}^{-2}$ )

  • Correct answer
    A
    $74 \mathrm{~ms}^{-2}$
  • B
    $55 \mathrm{~ms}^{-1}$
  • C
    $40 \mathrm{~ms}^{-1}$
  • D
    $60 \mathrm{~ms}^{-1}$

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, $\quad \mathrm{s}_1=\mathrm{s}_2$ $$ \begin{aligned} \frac{1}{2} g(18)^2 & =12 \mathrm{v}+\frac{1}{2} g(12)^2 \\ \Rightarrow \quad v & =74 \mathrm{~ms}^{-1} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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