At the same time as a stone is dropped from rest from a point A at the top of a 120m high cliff a ball is thrown vertically up with speed 30m/s from a point B which is at the bottom of the cliff and immediately below A. Find the height above B of the point where the ball and the stone collide.
Let , be the initial velocities of the ball and stone respectively.
Let , be the positions above ground of the ball and stone respectively.
Let and be acceleration and time from start respectively.
Then we have:
Then we have . We want :
you should be able to take it from there.
still confess to be honest
from your equation
at^2 on both side should cancel each other right?
which leave us with 120=30t
but when i put this in for x or s = ut+ 0.5*a*t^2
x therefore = 198
and yes it works for the other particle
but this is not what it's say in my answer on the back.
the answer should 41.6m
sorry for wasting your time!