FreeFall motion Assumptions in proving

[nabet94]

Homework Statement

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.

Homework Equations

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

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?

Homework Statement

initial velocity is u and that time after second stone is thrown is T

 

[Hypersphere]

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

 

[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 that's all i could get

 

[nabet94]

what do you mean with coordinate axis?

 

[Hypersphere]

what do you mean with coordinate axis?

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?

 

[nabet94]

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

 

[Hypersphere]

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

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.

 

[nabet94]

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

 

[Hypersphere]

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

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

 

[nabet94]

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

 

[Simon Bridge]

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