Question

A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from the ground with a velocity of `25m//s`. Calculate when and where the two stone will meet.

Answer 1

Here, `h=100m`.
Let the stones meet after t seconds at a point P which is at a height x above the ground as shown in

For stonees 1 `u=0,h=(100-x)m, a=g=9.8m//s^(2)`
From `s=ut+1/2at^(2)`,
`(100-x)=0+1/2xx9.8t^(2)=4.9t^(2)` ..(i)
For stone 2
`u=25m//s, h=x, a=-g=-9.8m//s^(2)`
From `s=ut+1/2at^(2), x=25t+1/2(9.8)t^(2)-4.9t2`
add eqns. (i) and (ii), `100-x+x+=25t,t=100/25=4s`
From eqn. (i), `100-x=4.9(4)^(2)=78.4`
`x=100-78.4=21.6m`