A ball is dropped from rest at t = 0. After 6 s, another ball is thrown downward from the same point with speed v. Both balls meet at t = 18text s . Find v (g = 10text m/s ^2).

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NEET Physics QUESTION #7314

Question 1

A ball is dropped from rest at $t = 0$. After 6 s, another ball is thrown downward from the same point with speed $v$. Both balls meet at $t = 18\text{ s}$. Find $v$ ($g = 10\text{ m/s}^2$).

60 m/s
75 m/s✔️
55 m/s
40 m/s

Correct Answer Explanation

Ball 1 falls for 18 s: $h_1 = \dfrac{1}{2}(10)(18)^2 = 1620\text{ m}$

Ball 2 falls for $18 - 6 = 12\text{ s}$: $h_2 = 12v + \dfrac{1}{2}(10)(12)^2 = 12v + 720$

Setting $h_1 = h_2$: $12v + 720 = 1620 \Rightarrow 12v = 900 \Rightarrow v = \mathbf{75\text{ m/s}}$

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