h first one: Vi*t-4.9t^2
h second one: Vi(t-4)-4.9(t-4)^2
because the heights are the same, set the equations equal, solve for t.
If both cans are thrown with the same velocity, then the first can must be on its way down when they collide, and the first can will be going up.
For the first can,
vt-4.9t^2 = 5
t = 5/49 (v+√(v^2-98))
the 2nd can has been going up for only t-4 seconds, so
v(5/49 (v+√(v^2-98))-4)-4.9((5/49 (v+√(v^2-98)))-4)^2 = 5
v = 21.958 m/s ... call it 22m/s
22t-4.9t^2 = 5 at t=4.25s and at t=0.25s
So, the 2nd can meets the 1st on its way down in 0.25 seconds.
Let the initial velocity be v.
At some time Δt the displacement of both cans is the same (h), so by using the the equations for displacement under constant acceleration, we obtain two simultaneous equations:
For the first can:
h = v Δt - g (Δt)^2 /2
For the second can (launched at t1):
h = v (Δt-t1) - g (Δt-t1)^2 /2
We need to find Δt
Rearrange the first equation:
v = h/Δt + gΔt/2
Substitute this into the second equation.
h = (h/Δt + gΔt/2)*(Δt-t1) - g*(Δt-t1)^2/2
This will simplify into a quadratic equation which can be solved for Δt.
Once you have t then use (from above):
v = h/t + gt/2
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