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An object is released from rest from the top of a tower of height 80 m. At the same instant another object is projected vertically upward with a velocity 30 m/s from the bottom of the tower along the same vertical line. Find the time after which the objects will collide.

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2.16 sec

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2.66 sec

3.16 sec

answer is A.

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Detailed Solution

If the particles meet after time t , then $\frac{1}{2} g t^{2} + (u t - \frac{1}{2} g t^{2}) = h \Rightarrow t = \frac{h}{u} = \frac{80}{30} s = 2.66 s$

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