Mary wants to throw a can straight up into the air and then hit it with a second can. She wants the collision to occur at height h=5.0 m above the throw point. In addition, she knows that she needs t1=4.0 s between successive throws. Assuming that she throws both cans with the same speed. Take g to be 9.81 m/s2.

Question

(a) How long it takes (in s) after the first can has been thrown into the air for the two cans to collide?

Δt=

Answer 1

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.

Answer 2

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

Check:
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.

Answer 3

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