Answer to Question #80685 in Classical Mechanics for Jithin

Question

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, s/he knows that she needs t1=3.0 s between successive throws. Assuming that s/he throws both cans with the same speed. Take g to be 9.81 m/s2.

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-t_1 )- g (t-t_1 )^2/2
x(t) = y(t) = h
From the first height equation:
v =h/t+gt/2
Substituted into the second equation:
h = (h/t+gt/2)(t-t_1 )-(g(t-t_1 )^2)/2
5 = (5/t+9.81 t/2)(t-3)-(9.81(t-3)^2)/2
t=3.308 s.
v =5/3.308+9.81/2 3.308=18 m/s.