Chapter 8, Problem 56SGA

To determine

To Find: The height of the ball is 0.5 second.

Answer to Problem 56SGA

The height of the ball is $\text{5}\text{.85}\text{ft}$ .

Explanation of Solution

Given:

The initial velocity of the ball is 38 feet per second.

The angel of the ball is $40°$ from the horizontal.

The ball is 2 feet above the ground.

Calculation:

Consider the horizontal and the vertical displacements of the projective at any time is,

  $\begin{array}{l}x=t\left|\overrightarrow{v}\right|\cos \theta \\ y=t\left|\overrightarrow{v}\right|\sin \theta -\frac{1}{2}g{t}^2\end{array}$

Then, from the given data.

  $\begin{array}{c}y=\left(0.5\right)\left(38\right)\sin \left(40°\right)-\frac{1}{2}\left(32\right){\left(0.4\right)}^2\\ =7.85-4\\ =3.85\text{ft}\end{array}$

Since, the ball was 2 ft above the ground then,

  $3.85\text{ft}+2\text{ft}=\text{5}\text{.85}\text{ft}$