2 balls one thrown upwards and one dropped?

First off...it's not actually 1/3 the height of the building that they meet at, it's at 2/3s of the height of the building (counting up from the ground that is).

If we start by defining the final velocity of ball B as our unit of velocity, then the final velocity of ball B is 1, and the final velocity of ball A is -2 (negative because it is going downwards). We know the initial velocity of ball A is 0, and because it is undergoing constant acceleration, its average velocity over the time of the experiment will be (initial + final) / 2, or -1.

Considering ball B, and knowing that all objects undergo the same acceleration due to gravity when air friction is ignored, we can conclude that its velocity has also decreased by two from its initial velocity, just as the velocity of ball A dropped from 0 to -2, so the initial velocity of ball B must have been 3 in order for it to still have a speed of 1 after the time interval of our experiment. So the average velocity of ball B over the experiment is (3 + 1) / 2 = 2.

Since we know the average velocity of ball A was -1, we can conclude it dropped a distance d = (average velocity) * time, or -1t (while units were left off in this example, it is always necessary to multiply velocity by time to get distance, e.g. 1 mile/hour * 1 hour = 1 mile). Over the same interval, Ball two rose the distance 2t. Since the height of the building = the distance ball B rose - the distance ball A fell, the building must be 3t in height, meaning that ball B rose 2/3s of the building height, and ball A fell 1/3 of the building's height.

Let t be the time they need to meet. then the dropped ball has the speed

vA = g * t

while the rising ball has the speed

vB = 1/2 vA = 1/2 gt

Ball A fell

f = 1/2 g t^2

Ball B rose

r = 1/4 gt^2

the heigth is

h = r + f = 3/4 gt^2 = 3 * r

So

h = 3 * r ---> r = 1/3 * h

I am fixing map's solution above me.

Let t = t_o be the time they need to meet. Then the dropped ball has the speed

vA = g * t_o

while the rising ball has speed

vB = u - g*t_o

We are given vB = 1/2 * v_A

substituting g * t_o for vA

we have vB = 1/2 * g*t_0

So far we have

vB = 1/2 * g*t_o

vB = u - g*t_o

By equality

1/2 * g*t_0 = u - g*t_o

solving for u

u = 3/2*g*t_0

Ball A fell

f = 1/2 g t_o^2

Ball B rose

r = u*t_0 - 1/2*gt_o^2

substitute u = 3/2*g*t_o

Ball B rose

r = (3/2*g*t_o) * t_o - 1/2*gt_o^2

r = g*t_o^2

the heigth (sic) is

h = r+f = g*t_o^2 + 1/2*g*t_o^2 = 3/2 gt_o^2 = 3/2 * r

So

h = 3/2 * r ---> r = 2/3 * h

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