Ball A is dropped from the top of a building of height h at the same time that Ball B is thrown vertically upward from the ground. When the balls collide, they are moving in opposite directions, and the speed of A is twice the speed of B. At what height does the collision occur?
xf=xi + 1/2(vi+vf)t
V(ball A) = 2V(ball B)
The Attempt at a Solution
I've done kinematics equations before, but it's been a while and there is a little more involved with this one, as you're looking for the moment of impact's height where the velocity of falling ball A is twice the speed of rising ball B.
I know the velocity of ball A is simply V(Ball A) = -9.8m/s^2*t
Ball B has a velocity of V(ball B) = V-9.8m/s^2*t
The collision must occur before or at the maximum height Ball B can reach with its initial velocity, but since I cant solve for it, I'm not sure what to do.
I'm fairly sure I can set the values equal to each other because of the relationship of V(ball A) to V(ball B) but I'm having trouble conceptualizing, and dont know if they even want an exact or symbolic answer.
V(ball A) = -9.8*t
V(ball B) = (-9.8*t)/2
I need to solve for time to get height, and height to get time. But I don't have any values for velocity. The only value I have is the constant of acceleration and I feel like that isn't enough to work this out.