Question

A can is thrown straight up into the air and then hit it with a second can. The collision to occur at height h=5.0 m above the throw point. In addition, that t1=4.0 s between successive throws. Assuming that both cans are thrown with the same speed. Take g to be 9.81 m/s.(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

setup equations and solve them

Answer 2

at the point of collision, both cans must be at same height : 5m

Answer 3

first can :
\(\large 5=ut-\frac{1}{2}9.81 t^2\)

second can :
\(\large 5=u(t-4)-\frac{1}{2} 9.81 (t-4)^2\)

Answer 4

solving should be simple, give it a try :)