# Homework Help: Basic kinematics : 1-Dimensional collision in mid-air

1. Aug 10, 2009

### Matty R

Hello.

I've been stuck on this question for ages, and was hoping someone could help me with it.

1. The problem statement, all variables and given/known data
A stone is projected vertically upwards with a speed of 6m/s, and one second later a second stone is projected vertically upwards from the same point with the same speed.

Find where the two stones meet.

2. Relevant equations

http://img193.imageshack.us/img193/4696/4bes.jpg [Broken]

3. The attempt at a solution

http://img195.imageshack.us/img195/908/4bp1.jpg [Broken]

http://img140.imageshack.us/img140/1691/4bp2.jpg [Broken]

http://img193.imageshack.us/img193/4383/4bp3.jpg [Broken]

http://img148.imageshack.us/img148/6464/4bp4.jpg [Broken]

As far as I can tell, I've worked out how high above the projection point the first stone is, when the second stone is launched.

I also worked out that the first stone is travelling at 4m/s at 1m above the ground after 1 second.

I tried doing this through trial and error, and came up with the collision at 0.15m above the projection point, 0.17 seconds after the second stone is launched, but I don't think that is the method I'm supposed to be using here. :shy:

I don't think I've ever done a question like this before, so I would really appreciate any help.

Thanks.

Last edited by a moderator: May 4, 2017
2. Aug 10, 2009

### boris1907

for first particle $$$s_1 = ut_1 - \frac{1}{2}gt_1^2$$$
for second particle $$$s_2 = ut_2 - \frac{1}{2}gt_2^2$$$
we know that $$$t_1-t_2=1 , t_2=t_1-1$$$
then $$$s_2 = u(t_1-1) - \frac{1}{2}g(t_1-1)^2$$$.
since they meet, s_1 should be equal to s_2.
$$$s_1 = ut_1 - \frac{1}{2}gt_1^2 = u(t_1-1) - \frac{1}{2}g(t_1-1)^2=s_2$$$
after solving the equation above for u=6m/s and g=10m/s^2, we find $$$t_1=1.1s ,t_2=0.1s$$$ and $$$s_1=s_2=0.55m$$$ .They meet at 0.55m above the ground.

3. Aug 11, 2009