Kinematics Question 408

Question: 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 = 18s. What is the value of v? $ (takeg=10m/s^{2}) $

Options:

A) 75 m/s

B) 55 m/s

C) 40 m/s

D) 60 m/s

Show Answer

Answer:

Correct Answer: A

Solution:

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

Now, distance moved in 18 s by 1st ball $ =\frac{1}{2}\times 10\times 18^{2}=90\times 18=1620\text{ m } $

distance moved in 12 s by 2nd ball $ =ut+\frac{1}{2}gt^{2}\text{ }\therefore \text{1620=12v+5}\times 144 $
$ \Rightarrow \text{v=135-60=75 m}{{s}^{-1}} $



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