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(D) Let $t$ be the time elapsed since body $A$ was projected. The displacement of body $A$ is given by $h_A = ut - \frac{1}{2}gt^2$.Since body $B$ is

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$12$

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(D) Let $t$ be the time elapsed since body $A$ was projected. The displacement of body $A$ is given by $h_A = ut - \frac{1}{2}gt^2$.Since body $B$ is projected $4 \, s$ later,its time of flight is $(t-4) \, s$. Its displacement is $h_B = u(t-4) - \frac{1}{2}g(t-4)^2$.When the bodies meet,$h_A = h_B$.$98t - \frac{1}{2} 	imes 9.8 	imes t^2 = 98(t-4) - \frac{1}{2} 	imes 9.8 	imes (t-4)^2$.$98t - 4.9t^2 = 98t - 392 - 4.9(t^2 - 8t + 16)$.$98t - 4.9t^2 = 98t - 392 - 4.9t^2 + 39.2t - 78.4$.$0 = -392 + 39.2t - 78.4$.$39.2t = 470.4$.$t = \frac{470.4}{39.2} = 12 \, s$.

$A$ body $A$ is projected upwards with a velocity of $98 \, m/s$. $A$ second body $B$ is projected upwards with the same initial velocity but after $4 \, s$. Both the bodies will meet after how many seconds (from the start of body $A$)?

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