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A body is thrown vertically up with a ve...

A body is thrown vertically up with a velocity of 100 m/s and another one is thrown 4 sec after the first one. How long after the first one is thrown will they meet?

Text Solution

Verified by Experts

Let them meet after t sec.
`S_(1)=100t-1/2gt^(2) and S_(2)=100(t-4) -1/2g(t-4)^(2)`
`100t-1/2gt^(2)=100(t-4)-1/2g(t-4)^(2)`
`400=1/2g[t^(2)-(t-4)^(2)]=1/2 g.4 (2t-4)`
`2t-4=(800)/(4g)=20," if g =10m/"s^(2)`
t=12sec
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