Basic kinematics : 1-Dimensional collision in mid-air

[Matty R]

Hello.

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

Homework Statement

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.

Homework Equations

http://img193.imageshack.us/img193/4696/4bes.jpg

The Attempt at a Solution

http://img195.imageshack.us/img195/908/4bp1.jpg

http://img140.imageshack.us/img140/1691/4bp2.jpg

http://img193.imageshack.us/img193/4383/4bp3.jpg

http://img148.imageshack.us/img148/6464/4bp4.jpg

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 traveling 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.

 

[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.

 

[Matty R]

Thanks for the reply.

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.

Thank you so much for this. It's exactly what I was after. I haven't done anything like this before, so it took some time to get my head around it, but I think I understand it all now. I managed to work through it all and end up with 0.55 for s1 and s2.

Thanks again.