Question:
A body is projected vertically up at t=0 with a velocity of 98m/s. Another body is projected from the same point with same velocity after time 4 seconds. Both bodies will meet after how many seconds
Answer:
Let us consider that "t" is be the time of flight of the first body when it meets the second body, so the time of flight of the second body at which it meets the first body in terms of "t" is equal to "t-4".
The distance or max height reached by both is the same, so putting the formula S= ut + 1/2 gt2 for both bodies should be equal.
That is, 98t - 1/2 x g x t2 = 98(t-4) - 1/2 x g x (t-4)2 [Here g = -a]
Solving it, 98t - 98(t-4) = 1/2 x gt2 - 1/2 g(t-4)2
So, 98[t - (t-4)] = 1/2 x 9.8 x [t2 - (t-4)2]
Therefore, 20 [t - t - 4] = [t2 - t2 + 8t - 16]
That is, 80 = 8t - 16
Therefore, 8t = 96
t = 12 seconds.
Thus, the bodies will meet after 12 seconds.
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