JEE MAIN - Physics (2022 - 26th June Evening Shift - No. 18)
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
6
Explanation
At t = 2 s, v1 = 50 $$-$$ 2 $$\times$$ 10 = 30 m/s
v2 = v2
$$\therefore$$ arel = g $$-$$ g = 0
$$S = {{{u^2} - {v^2}} \over {2g}} = {{{{50}^2} - {{30}^2}} \over {2 \times 10}} = {{1600} \over {20}} = 80$$ m
$$\therefore$$ vrel = 50 $$-$$ 30 = 20 m/s
$$\therefore$$ $$\Delta t = {{80} \over {20}} = 4\,s$$
$$\therefore$$ required time t = 2 + 4 = 6 s
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