Q 11-02-095JEE MainJEE Main 2022 (26 Jun, Shift 2)Easy
A ball is projected vertically upward with an initial velocity of $50\ \text{m s}^{-1}$ at $t = 0\ \text{s}$. At $t = 2\ \text{s}$, another ball is projected vertically upward with the same velocity. At $t$ = ______ s, the second ball will meet the first ball. ($g = 10\ \text{m s}^{-2}$)
Numerical value type. Enter your answer.
Answer: 6
$$50t - 5t^2 = 50(t-2) - 5(t-2)^2$$
$$0 = -100 - 5(-4t + 4)\ \Rightarrow\ 20t = 120\ \Rightarrow\ t = 6\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics