A ball is projected vertically upward with an initial velocity of 50 ms$-$1 at t = 0s. At t = 2s, another ball is projected vertically upward with same velocity. At t = __________ s, second ball will meet the first ball (g = 10 ms$-$2).
Answer (integer)
6
Solution
<p>At t = 2 s, v<sub>1</sub> = 50 $-$ 2 $\times$ 10 = 30 m/s</p>
<p>v<sub>2</sub> = v<sub>2</sub></p>
<p>$\therefore$ a<sub>rel</sub> = g $-$ g = 0</p>
<p>$$S = {{{u^2} - {v^2}} \over {2g}} = {{{{50}^2} - {{30}^2}} \over {2 \times 10}} = {{1600} \over {20}} = 80$$ m</p>
<p>$\therefore$ v<sub>rel</sub> = 50 $-$ 30 = 20 m/s</p>
<p>$\therefore$ $\Delta t = {{80} \over {20}} = 4\,s$</p>
<p>$\therefore$ required time t = 2 + 4 = 6 s</p>
About this question
Subject: Physics · Chapter: Kinematics · Topic: Motion in a Straight Line
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