Question

A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 s 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{{s}^{-2}} $ )

• A. $ 74\,m{{s}^{-1}} $
• B. $ 64\,m{{s}^{-1}} $
• C. $ 84\,m{{s}^{-1}} $
• D. $ 94\,m{{s}^{-1}} $

Answer 1

Answer: (A)

For first ball, $ u=0 $ $ \therefore $ $ {{s}_{1}}=\frac{1}{2}gt_{1}^{2}=\frac{1}{2}\times g{{(18)}^{2}} $ For second ball, initial velocity $ =v $ $ \therefore $ $ {{s}_{2}}=v{{t}_{2}}+\frac{1}{2}\times g{{t}^{2}} $ $ {{t}_{2}}=18-6=12\,s $ $ \Rightarrow $ $ {{s}_{2}}=v\times 12+\frac{1}{2}g{{(12)}^{2}} $ Here, $ {{s}_{1}}={{s}_{2}} $ $ \frac{1}{2}g{{(18)}^{2}}=12v+\frac{1}{2}g{{(12)}^{2}} $ $ \Rightarrow $ $ v=74\,m{{s}^{-1}} $