A stone is dropped from a height of 100 m and at the same time another stone is thrown vertically upwards with velocity of `40ms^(-1)`. When and where will the two stones meet ? `(g=10ms^(-2))`

Explanation

Given `h=100m, v_(0)=40ms^(-1)`
`g=10ms^(-2)`
Let the two stones meet at B at a height of `(100-x)` from the ground. For the I particle and for the II particle, `(100-x)=v_(0)t-1//2g t^(2)`
`"i.e "100-1//2g t^(2)=v_(0)t-1//2g t^(2)`
`therefore" "t=(100)/(40)=2.5s" Hence "x=(1)/(2)xx10xx2.5xx2.5`
Hence `x=(1)/(2)xx10xx2.5xx2.5=5xx6.25`
`x=31.25m`
The two stone will meet at hight 68.75 m above the ground.