Question

A body is projected vertically up at t=0 with a velocity of 98m/s. after 2s if the acceleration

Answer 1

answer is
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.

Answer 2

Answer:

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