Question

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?

Short answer

The roof’s height is 14.5 m.

Step-by-step solution

Step 1. Assumptions for the motion of the ball

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.

Step 2. Identification of the given data

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 $y=0$.

Step 3. Determination of the time of flight during the horizontal motion

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,

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

Step 4. Determination of the height of the building

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.