ExamBro
ExamBro
JEE Mains · Physics · STD 11 - Motion in straight line
  • JEE Mains 2022
  • Hard
  • Numerical

Two balls \(A\) and \(B\) are placed at the top of \(180 \,m\) tall tower. Ball \(A\) is released from the top at \(t =0 \,s\). Ball \(B\) is thrown vertically down with an initial velocity \(u\)' at \(t=2\, s\). After a certain time, both balls meet \(100 \,m\) above the ground. Find the value of \(u\) in ............... \(ms ^{-1}\). [use \(g =10 \,ms ^{-2}\) ]

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(D) \(30\)

Step-by-step Solution

Detailed explanation

Let they meet at time \(t\).

\(t =\sqrt{\frac{2 h }{ g }}=\sqrt{\frac{2 \times 80}{10}}\)

\(=4 \,sec\)

Time taken by ball \(B\) to meet \(A =2 \,sec\)

\(\text { using } S=u t+\frac{1}{2} \text { at }^{2}\)

\(-80=-u \times 2+\frac{1}{2}(-10)(2)^{2}\)

\(u=30\)

Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app