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.
many thanks!
Okay, you given it a fair shot, let's see ...
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
therefore t=4
but when i put this in for x or s = ut+ 0.5*a*t^2
when
u=30
a=-9.81
t=4
x therefore = 198
and yes it works for the other particle
120+78=198
but this is not what it's say in my answer on the back.
the answer should 41.6m
many thanks!
sorry for wasting your time!