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?

(b) Find the initial speed of the cans (in m/s).

Answer 1

Let the height of the first can be x, that of the second can be y; and both cans be thrown at speed v.

x(t) = v t - g t^2 /2
y(t) = v (t-t1) - g (t-t1)^2 /2

Find t when x(t) = y(t) = h

From the first height equation:
v = h/t + gt/2

Substituted into the second.
h = (h/t + gt/2)(t-t1) - g(t-t1)^2/2

Simplify into a quadratic and solve for t.

...

Once you have t then use (from above):
v = h/t + gt/2

Answer 2

thank you it help me!