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Question

A stone is dropped from a height h simultaneously another stone is thrown up from the ground which reaches the height 4h.The two stones cross each other after a time.

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Solution

u is the velocity with which second stone is projected .v2-u2 = 2asat heighest point v = 0s= 4ha= -g0 -u2 = -2g4hu=8ghlet t is the time when they cross eath other.in t time distance tarvel by first stone is:x = ut+0.5gt2x = 0+0.5×10t2x = 5t2distance travelled by second stone is :h-x = ut-5t2h -5t2 =8ght - 5t2h = t8ght = h8g

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