Stone dropped from top of a tower 100m high. At same instant another stone is thrown vertically from base of the tower with a velocity of 25m/s. When and where will the two stones meet? Given g =10 m/s^2

Answer 1

see other post.

Answer 2

let the height of the dropped stone be s1
s1 = ut + 0.5gt^2
u=0
s1 = 0.5*10*t^2 = 5t^2

let the height of the thrown stone be s2
s2 = ut - 0.5gt^2
s2 = 25t - 5t^2

s1+s2 = 100
5t^2 + 25t - 5t^2 = 100
25t = 100
t = 4s

the stones meet 4seconds later

s2 = 25t - 5t^2 = 100 - 5(16)
s2 = 100 - 80 = 20m

the stones meet 20m from the base of the tower.

Answer 3

Thanks helped me a lot

Answer 4

very helpful. thanx a lot