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Kinematics

JEE Main level, single correct, medium

164 of 308

Question 164 of 308

Two balls $A$ and $B$ are placed at the top of $180m$ tall tower. Ball $A$ is released from the top at $t=0s$ . Ball $B$ is thrown vertically down with an initial velocity $u$ at $t=2s$ . After a certain time, both balls meet $100m$ above the ground. Find the value of $u$ in $m{s}^{-1}$ . [use $g=10{ms}^{-2}$ ]

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Answer: (D) $30$

Displacement covered by first ball,

\left(180-100\right)=0+\dfrac{1}{2}\times 10\times {t}^{2}

\Rightarrow t=4s

Now, the second body gets only,

{t}_{2}=4-2=2s

.
Displacement covered by second will be same,

80=u\times 2+\dfrac{1}{2}\times 10\times {\left(2\right)}^{2}

\Rightarrow u=\dfrac{80-20}{2}=30m{s}^{-1}

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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