JEE Mains · Physics · STD 11 - 2. motion in straight line
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 \(g =10 \,ms ^{-2}\) ]
- A \(10\)
- B \(15\)
- C \(20\)
- D \(30\)
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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