Two bodies being a free fall from rest, from the same height 2s apart. How long after the first body begins to fall the two bodies will be 40m apart? (Take $g = 10ms^{- 2}$)

Question

Dr. Abubacker · Accepted Answer

(c) $\frac{1}{2}gt^{2} - \frac{1}{2}g(t - 2)^{2} = 40$
or $\frac{1}{2} 	imes 10(2t - 2)(2) = 40$ or $2t - 2 = 4$ or $t = 3$s

Q. Two bodies being a free fall from rest, from the same height 2s apart. How long after the first body begins to fall the two bodies will be 40m apart? (Take $g = 10ms^{- 2}$)

Explanation

Reviewed by Dr. Abubacker

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