A ball thrown horizontally at$12.2\mathrm{m}/\mathrm{s}$from the roof of a building lands 21.0 m from the base of the building. How high is the building?

Suppose the positive y-direction is downward. Consider the origin as the point from where the ball is thrown from the roof of the building.

Along the horizontal direction:

The initial horizontal velocity is $v_{\mathrm{x}}=12.2\mathrm{m}/\mathrm{s}$.

The horizontal range is $x=21\mathrm{m}$.

Along the vertical direction:

The initial velocity is $v_{\mathrm{y}}=0$.

The acceleration experienced by the ball is $a_{\mathrm{y}}=g=9.8\mathrm{m}/{\mathrm{s}}^2$.

The vertical distance traveled by the ball is $y=0$.

The ball is traveling at a constant velocity during its horizontal motion. The time taken by the ball is given as the distance divided by the velocity. Mathematically,

$\begin{array}{c}t=\frac{x}{v_{\mathrm{x}}}\\ =\frac{21\mathrm{m}}{12.2\mathrm{m}/\mathrm{s}}\\ =1.72\mathrm{s}\end{array}$

The vertical displacement is equivalent to the height of the building, h. Using the kinematic equation of motion along the vertical direction, you get:

$\begin{array}{c}y=h=v_{\mathrm{y}}t+\frac{1}{2}g{t}^2\\ h=0+\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(1.72\mathrm{s}\right)}^2\\ =14.5\mathrm{m}\end{array}$

Thus, the height of the building is 14.5 m.