Question

A stone is drop from a height 19.6m above the ground while a second stone is simultaneously projected from the ground with sufficient velocity to enable it to accept 19.6m. When and where the stone will meet

Answer 1

dropped stone:
hf=hi+vi*t-4.9t^2
hf=19.6-4.9t^2
thrown stone:
Initial KE =final PE of max height
1/2 mvi^2=mg*19.6
vi=sqrt(2*9.8*19.6)
vi= 19.6m/s
hf=hi+vi*t-4.9t^2
hf=19.6*t-4.9t^2
but the hf is the same for each stone.
19.6*t-4.9t^2 =19.6-4.9t^2
they meet at time t=1second
hf=19.6*t-4.9t^2....