Question
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
Answers
bobpursley
see other post.
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.
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.
Thanks helped me a lot
very helpful. thanx a lot
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