Chapter 3, Problem 23P

To determine

The height of the building.

Answer to Problem 23P

12.8 m

Explanation of Solution

Given:

The initial velocity of ball is 22.2 m/s, distance between base of building and ball lands is 36 m.

Formula Used:

The kinematics equations are,

  $\begin{array}{l}{u}_x=\frac{S_x}{t}\\ S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2\end{array}$

Calculation :

  $\begin{array}{l}\text{Initial velocity, }{u}_x=22.2\text{ m/s}\\ \text{Horizontal distance, }{S}_x=36\text{ m}\end{array}$

Substituting the values, to find the time.

  $\begin{array}{l}22.2=\frac{36}{t}\\ t=\frac{36}{22.2}\\ t=1.6\text{ sec}\end{array}$

Displacement along y axis is,

  $S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

Substituting the given values

  $\begin{array}{l}{S}_y=\frac{1}{2}\times 10\times {\left(1.6\right)}^2\\ S_y=12.8\text{ m}\end{array}$

Conclusion:

Thus, the height of building is 12.8 m.