Two balls A and B are placed at top of 180 ,m tall tower. Ball A is released from top at t =0 ,s . B

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2.Motion in Straight Line

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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

$30$

(JEE MAIN-2022)

Solution

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 $A =2 \,sec$

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

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

$u=30$

Standard 11

Physics

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