Mechanics..Projectile?

Question

A particle is projected vertically upwards from a fixed point O with speed u. A short time T later a second particle is similarly projected from O. prove that the particles will collide at time t after the first particle is projected where

t= 1/2 T + u/g

Show that the speeds of the two particles immediately before impact are equal.

I have gone around and around but dont seem to get any where, any help greatfully received.

Answer 1

For each particle, the acceleration is a = -g
dv/dt = -g
v = -gt + u
s = -gt^2 / 2 + ut
This is for particle 1.

For the second particle, the only difference is that you start at t = T, so you just have to translate.
s2 = -g*(t-T)^2 / 2 + u*(t-T)

Now we set them equal at the time they collide:
-gt^2 / 2 + ut = -g*(t-T)^2 / 2 + u*(t-T)
ut - u*(t-T)= gt^2 / 2 -g*(t-T)^2 / 2
u*T = g/2 *[2t - T]*T
u = g/2 *(2t - T)
2t - T = 2u /g
t = u/g + T/2
---
Now for the speeds.
v1 = -gt + u
v2 = -g(t-T) + u
We wish to prove that their SPEEDS are equal. Obviously they have opposite directions.
So consider v1 + v2 =
-g*(2t+T) + 2u
= 0 as you can see.

So their speeds are equal, directions opposite.

Answer 2

The equation for a projectile shot upward is
height = 0.5*g*t^2 + v * t, where g is the gravitional accelertion, v is the orginal velocity (u), so for the first particle height(1) = 0.5*g*t(1)^2 + U * t(1) and for the
second particle height(2) = 0.5*g*t(2)^2 + u*t(2). where
the t(1) and t(2) is the elapsed time from firing (launched)

to Collide height(1) = height(2), I.e. particle one is on the way back down. therefore
0.5 * g*t(1)^2 + u*t(1) = 0.5*g*t(2)^2 + u * t(2)
but t2 = t1 - T (second particle was fired later). then
0.5 * g*t(1)^2 + u*t(1) = 0.5 * g * (t(1)-T)^2 = u*(t(1)-T)
upon squaring (t(1)-T) = t(1)^2 - 2*t(1)*T + T^2
substituting that in, we have
0.5 * g * t(1)^2 + u*t(1) = 0.5 * g* (t(1)^2-2*t(1)*T+T^2) + u*(t(1)_T)
or
0.5*g*T(1)^2 +u*t(1)=0.5*g*t(1)^2 -g*t(1)*T+0.5*g*T^2+U(t1) - u*T,
cancelling out terms,
0 = -g*t(1)*T + 0.5*g*T^2 - u*T
dividing out T
0 = -g*t(1) + 0.5*g*T - u or
g*t(1) = 0.5*g*T - u or
t(1) = 0.5*T - u/g
I think there is a sign error in your answer.

Answer 3

It's simple. Conservation of energy requires that both masses have the same speed when they collide. But the they have opposite velocities - that is the first mass is traveling downward at some speed v, while the second mass is travelling up at v. So the speed of the first mass can be written as:

v1 = -g*t + u

and the second mass as

v2 = -g*(t-T)+u

Now since v1 = -v2:

gt-u = -g(t-T)+u

Re-arranging:

2gt = gT +2u

So you get

t = 1/2*T + u/g

:>)