Q.From the terrace of a building of height H, you dropped a ball of mass m. It reached the ground with speed v. Is the relation $mgH = \frac{1}{2}mv^2$ applicable exactly? If not, how can you account for the difference? Will the ball bounce to the same height from where it was dropped?

[!TLDR] Exactly only if air resistance is negligible; in reality some energy is lost to air drag, so the ball's actual landing speed is slightly less than the ideal $v=\sqrt{2gH}$, and it will NOT bounce back to the same height (some energy is also lost in the bounce itself). [!ANSWER] Only …

The relation $mgH=\frac{1}{2}mv^2$ is the work-energy theorem applied to a purely conservative (gravity-only) fall, i.e. it assumes NO air resistance. In that idealised case it is exactly true: all of the gravitational potential energy lost converts entirely into kinetic energy. In REALITY, air resistance (a non-conservative, path-dependent force) also acts during the fall, so the correct relation (4.6, Case II) is $mgH=\frac{1}{2}mv^2+W_{\text{air resistance}}$ -- some of the potential energy is used up overcoming drag and is dissipated as heat/sound, meaning the ball's ACTUAL landing speed is slightly LESS than the ideal $\sqrt{2gH}$ value the simple formula would predict. On top of this, even upon striking the ground the ball will not bounce back to exactly the same height H, because the ball-ground collision is itself never perfectly elastic (real coefficient of restitution $e<1$, 4.8.3) -- additional kinetic energy is lost as heat/sound/deformation at each bounce, so the rebound height is always somew …

Apply the work-energy theorem including the non-conservative air-resistance term, and separately note that a real …

Treating the two losses (air resistance during the fall, and imperfect elasticity of the bounce itself) as the same effect, rather than …