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(A) Let the two balls meet at time $t$ seconds from the start of the motion of the first ball.The displacement of the first ball at time $t$ is given

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$6$

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(A) Let the two balls meet at time $t$ seconds from the start of the motion of the first ball.The displacement of the first ball at time $t$ is given by $h_1 = 50t - \frac{1}{2}gt^2$.The second ball is projected at $t = 2 \; s$,so its time of flight is $(t - 2) \; s$. Its displacement is $h_2 = 50(t - 2) - \frac{1}{2}g(t - 2)^2$.Since they meet at the same height,$h_1 = h_2$.$50t - \frac{1}{2}gt^2 = 50(t - 2) - \frac{1}{2}g(t - 2)^2$$50t - 5t^2 = 50t - 100 - 5(t^2 - 4t + 4)$$50t - 5t^2 = 50t - 100 - 5t^2 + 20t - 20$$0 = -120 + 20t$$20t = 120$$t = 6 \; s$.

$A$ ball is projected vertically upward with an initial velocity of $50 \; ms^{-1}$ at $t = 0 \; s$. At $t = 2 \; s$,another ball is projected vertically upward with the same velocity. At $t = \dots \; s$,the second ball will meet the first ball $(g = 10 \; ms^{-2})$.

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