# Math Help - Kinematics: The Motion of Particles

1. ## Kinematics: The Motion of Particles

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!

2. Originally Posted by BabyMilo
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!
Both of these are constant-acceleration questions.

You want an expression for height in terms of time for both items. I'd be tempted to try $s = ut + \frac {at^2}2$ and see what happens.

3. would it be 30t+0.5*-9.8t^2=0t+0.5*-9.8t^2?

4. Originally Posted by BabyMilo
would it be 30t+0.5*-9.8t^2=0t+0.5*-9.8t^2?
Not quite, because one starts at the ground, one at 120m up. So the RHS should be (I think) 120 minus all that.

That gives you the time at which they hit. Then you just need to plug that back in to get the distance at they reach at that time.

5. 30t-4.9t^2=120+4.9t^2?
still cant work it out
can you please show the full working.

6. Originally Posted by BabyMilo
30t-4.9t^2=120+4.9t^2?
still cant work it out
can you please show the full working.
Okay, you given it a fair shot, let's see ...

Let $u_1$, $u_2$ be the initial velocities of the ball and stone respectively.

Let $s_1$, $s_2$ be the positions above ground of the ball and stone respectively.

Let $a$ and $t$ be acceleration and time from start respectively.

Then we have:

$s_1 = 120 + u_1 t - \frac {a t^2} 2$

$s_2 = u_2 t - \frac {a t^2} 2$

Then we have $u_1 = 0, u_2 = 30$. We want $s_1 = 120 + s_2$:

$30 t - \frac {a t^2} 2 = - \frac {a t^2} 2 + 120$

you should be able to take it from there.

7. still confess to be honest

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.

many thanks!

8. Originally Posted by BabyMilo
still confess to be honest

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.

many thanks!

$a$ is minus, yeah? You've added.

You're all right. This stuff is confusing in the early stages. (You're not wasting my time, I'm babysitting a program running on my other computer.)

9. got it thanks!