A ball is thrown vertically upwards.....?

For vertical projectiles, the equation governing the movement is:

y = -1/2gt² + V0t + y0

where y0 is the initial location of the projectile, and v0 its initial velocity.

Let's take the bottom of the building to be the origin for the problem.

1-> ball thrown downward, v0 = 0 and y0 = 100

2-> ball thrown upward, v0 = 25 and y0 = 0

Assuming that g = 10 m/s², we plug the inital values into the equation:

1->downward ball: y = -5t² + 100

2->upward ball: y = -5t² + 25t

If the balls collide, it means they have the same altitude (y), which means:

-5t² + 100 = -5t² + 25t

25t = 100

t = 100/25 = 4 seconds

which means, the two balls will collide after 4 seconds from being thrown.

Collided, so 100 m - s downward = s upward

100 m - Vo.t + 0,5.g.t^2 = Vo.t - 0,5.g.t^2

100 m - 0.t + 0,5.10.t^2 = 25.t - 0,5.10.t^2

100 m - 5t^2 = 25t - 5t^2

100 m = 25t - 5t^2 + 5t^2

100 m = 25t

t = 4s. They will collide after 4 seconds.

Here's another way to check.

DOWNWARD BALL

s = Vo.t + 0,5.g.t^2

s = 0.4 + 0,5.10.16

s = 0 + 80

s = 80 m

UPWARD BALL

s = Vo.t - 0,5.g.t^2

s = 25.4 - 0,5.10.16

s = 100-80

s = 20 m

Downward + Upward = 80 m + 20 m = 100 m

So, it's true, 4 seconds.

1/2 A T^2 = distance, so 3 sec = 45 m &(30 m/s) down, up at 10m/s should stop in one meter with 5 meter displ.