# FreeFall motion Assumptions in proving

1. Oct 10, 2012

### nabet94

1. The problem statement, all variables and given/known data
a stone is thrown vertically upwards with a speed of u metres per second. a second stone is thrown vertically uppwards from the same point with the same initial speed but T seconds later than the first. prove that they collide at a distance (4u^2-g^2T^2)/(8g) metres above the point of projection.

2. Relevant equations
s=ut+.5(g)(t)^2

3. The attempt at a solution
now i should resolve vertically so i would get s= uT + .5(g)T^2 for the second and s= u(X-T)+ .5(g)(X-T)^2 for the first but when i attempt to get the answer i fail miserably or everything just cancels out. can someone help me with this?
1. The problem statement, all variables and given/known data
initial velocity is u and that time after second stone is thrown is T

2. Oct 10, 2012

### Hypersphere

Can you show us what you have done so far? And which way is your coordinate axis pointing?

3. Oct 10, 2012

### nabet94

well i got u(X-T)+ .5(g)(X-T)^2= uT + .5(g)T^2 thus it is equal to uX-2uT+.5gx^2-.5gXT thats all i could get

4. Oct 10, 2012

### nabet94

what do you mean with coordinate axis?

5. Oct 10, 2012

### Hypersphere

I mean, in which direction does the y axis point? Does it point upward? I.e. does higher positions correspond to higher values of y? Because, in that case the formula the gravitational acceleration is opposite to the y axis, and the acceleration $a=-g$. You would then have
$y_1 = ut - \frac{gt^2}{2}$
for the first stone, where $t$ is the time, and we let $y=0$ at the starting point of the throw. The other stone satisfies
$y_2 = u (t-T) - \frac{g}{2} \left( t-T \right)^2$.

OK, so then you say they collide when $y_1=y_2$. That's good, but what does it tell you?

6. Oct 10, 2012

### nabet94

oh yeah well its going upwards so yeah. well that the y's are the same. thats all i am able to assume

7. Oct 10, 2012

### Hypersphere

My point is the equaton $y_1=y_2$ does not tell you at which height they collide. It does, however, provide some other useful information.

8. Oct 10, 2012

### nabet94

such as the time it takes for them to collide? and their initial velocity perhaps?

9. Oct 10, 2012

### Hypersphere

One of them, certainly. The initial velocity is already known though, as it is given in the problem description.

10. Oct 10, 2012

### nabet94

i got that T=(2gt+1)/t

11. Oct 11, 2012

### Simon Bridge

Have you mixed up the roles of T and t?
(Do you remember which time, t or T, you need?)
Have you cancelled too many terms?