Question

Ball A is dropped from the top of a building of height h at the same instant 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? (Use any variable or symbol stated above as necessary.)

Answer 1

so it goes like this...
We denote the height of the building as h, the position of ball A as yA, the initial position of ball A as yA0, and analogously for ball B.

The equation of motion for ball A is the following:
yA - yA0 = vA0t - (0.5)gt2
yA0 = h
vA0 = 0
g = 9.81 m/s2
Therefore, yA = -(0.5)(9.81 m/s2)t2 + h
=> yA = -4.905t2 + h

The equation of motion for ball B is the following:
yB - yB0 = vB0t - (0.5)gt2
yB0 = 0
g = 9.81 m/s2
Therefore, yB = vB0t - (0.5)(9.81 m/s2)t2
=> yB = vB0t - 4.905t2

The two balls collide when yA = yB:

yA = yB => -4.905t2 + h = vB0t - 4.905t2
=> h = vB0t

For ball A:
vA = vA0 - gt
vA0 = 0
Therefore, vA = - gt => vA = -9.81t

For ball B:
vB = vB0 - gt
Therefore, vB = vB0 - gt => vB = vB0 - 9.81t

We also know that when yA = yB, vA = -2vB:

vA = -2vB => -9.81t = -2[vB0 - 9.81t]
=> -9.81t = -2vB0 + 19.62t
=> 2vB0 = 29.43t
=> vB0 = 14.715t

Since h = vB0t, we have:

h = (14.715t)t = 14.715t2

Unfortunately, I couldn't figure out what to do from this point until I spent the better part of the last hour typing this thread. Then, as I was typing the previous sentence, it hit me:

xA = -4.905t2 + h = -4.905t2 + 14.715t2
=> xA = 9.81t2

Therefore, xA / h = 9.81t2 / (14.715t2)
=> xA / h = 2/3
=> xA = 2h/3, Hope it helps.