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Motion in A Straight Line question

2010 · Q166
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Motion in A Straight Line question

2010 · Q166

NEETPhysicsMotion in A Straight LineMCQ+4 / −1
A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 seconds 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 = 10 m/s2)
  1. A
    75 m/s
  2. B
    55 m/s
  3. C
    40 m/s
  4. D
    60 m/s
View written solutionFree

Correct answer: A

From the question, we can say

distance moved by 1st ball in 18 s = distance moved by 2nd ball in 12 s.

So, distance moved by 1st ball in 18 s

= 12×10×182{1 \over 2} \times 10 \times {18^2}21​×10×182 = 1620 m

and distance moved by 2nd ball in 12 s

= vt+12gt2vt + {1 \over 2}g{t^2}vt+21​gt2

∴\therefore∴ 1620 = v(12)+12×10(12)2v\left( {12} \right) + {1 \over 2} \times 10{\left( {12} \right)^2}v(12)+21​×10(12)2

⇒\Rightarrow⇒ vvv = 75 m/s

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