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Question

A stone is thrown vertically upwards with a velocity odf 19.6m/s. After 2s another stone is thrown upwards with a velocity of 9.8m/s. When and where will the two stones collide??

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Solution

For the first stone,initial velocity = 19.6 m/stime taken to reach heighest point is tv = u + at0 = 19.6 - 9.8 tt = 2 smaximum height hh = ut + 12at2h = 19.6×2 - 0.5×9.8×4h=19.6 mwhen first stone is at the top most point then second stone is thrown.let x is the distance travelled where they meet in time t1x = 12gt12for second stone distance travelled is h-x h-x = 9.8t1+0.5×9.8×t12on solving these two equations we gett1 = 2 sand x = 19.6 mso these two stones meet at ground after 2+2 = 4 s.

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