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.
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 \(A =2 \,sec\)
\(\text { using } S=u t+\frac{1}{2} \text { at }^{2}\)
\(-80=-u \times 2+\frac{1}{2}(-10)(2)^{2}\)
\(u=30\)
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