Question about physics free fall and acceleration -?

The going up part is symmetrical with part of the going down part. Giving the parameters the way they were wipes out the rise and part of the fall.

During the trip, the acceleration on the ball (gravity) is the same on the ball (down at 9.8 m/s, etc) all the time. Once the ball has left the thrown hand, no other acceleration applies, so the ball starts slowing down until it reaches the peak, has zero upward velocity and starts down.

But since is could be moving at almost any velocity when it get back to the level of the top of the building, how to account for that? If thrown really high, it would be going very fast but if just barely tossed sideways to clear the building, its start would be really slow.

But that is taken care of in the very exact 59.4 meters of height and the 4 seconds after being thrown. In the extreme from just the rim of the building, one time and velocity will apply, but it won't be 4 seconds. So for a non-plug and chug method you would have to compute the gain in velocity for the 59.4m plus the initial velocity, which you can do in one formula.

Many problems are easier to understand in terms of energy.

The final kinetic energy is equal to the initial kinetic energy + the change in potential energy (mgh). So all that matters is the difference between initial and final PE.

As the ball goes up, its KE is converted to PE until at the top of the motion the KE is 0 and it's all PE. So that top height d (relative to the start point) is such that mgd = initial KE.

As it passes back past its original point, all of THAT PE has gone back to KE, and its KE is now exactly what it was to begin with. That is, it is moving as fast going downward as it was going upward. And then it gains more energy mgh where h is the height of the building.

Here's another way to look at it: The ball goes through these points in its path. (a) initial height h with upward velocity 59.4 m/s, (b) maximum height h_max with upward velocity 0, (c) height h with downward velocity 59.4 m/s.

So if you start it in state (b) or (c) you should get the same final answer as starting at (a), since it's going to go through states (b) and (c).

And a 3rd way to look at it (sort of the same as the 1st, but algebraically):

Final KE = initial KE + mgh. That's conservation of energy.

(1/2)mv_f^2 = (1/2)mv_i^2 + mgh

mv_f^2 = mv_i^2 + 2mgh

v_f^2 = v_i^2 + 2gh

The reason it incorporates both the going up and going down is that kinetic energy doesn't have a direction. This equation is true no matter what direction the initial motion had. Kinetic energy doesn't have a direction.

So in fact you can apply it to a projectile with partly horizontal motion, and use vi and vf to mean the total velocity, not just the vertical velocity.