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NEET 2010
Exam
NEET
Year
2010
Subject
Physics
Chapter
Motion in One Dimension

NEET 2010 · Physics · Motion in One Dimension

A ball is dropped from a high rise platform at t=0 starting from...

A ball is dropped from a high rise platform at t=0\mathrm{t}=0 starting from rest. After 6 s6 \mathrm{~s} another ball is thrown downwards from the same platform with a speed v\mathrm{v}. The two balls meet at t=18 s\mathrm{t}=18 \mathrm{~s}. What is the value of v\mathrm{v} ? (take g=10 ms−2g=10 \mathrm{~ms}^{-2} )

Options

  1. A74 ms−274 \mathrm{~ms}^{-2}
  2. B55 ms−155 \mathrm{~ms}^{-1}
  3. C40 ms−140 \mathrm{~ms}^{-1}
  4. D60 ms−160 \mathrm{~ms}^{-1}

Correct answer

A. 74 ms−274 \mathrm{~ms}^{-2}

Step-by-step solution

. For first ball, u=0\mathrm{u}=0 \therefore \quad s1=12gt1\mathrm{s}_{1}=\frac{1}{2} \mathrm{gt}_{1}^2=12×g(\frac{1}{2} \times \mathrm{g}(18)^2 For second ball, initial velocity =v=\mathrm{v}∴s2=vt2+12gt2⇒t2=18−6=12 s⇒s2=v×12+12 g(12)2\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} \times₁₂⁺\frac{1}{2} \mathrm{~g}(12)^2 \end{aligned}Here,Here,\quad \mathrm{s}_1=\mathrm{s}_2 12g(18)2=12v+12g(12)2⇒v=74 ms−1\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}
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