Chapter 9.3, Problem 28PPS

a.

To determine

To find the duration of time that the ball is in the air.

Answer to Problem 28PPS

The ball is in the air for 3 seconds.

Explanation of Solution

Given information :

The function provided for the path of the ball is $h=-16t^2+47t+3$ .

Formula used :

The given equation would be graphed as a function and the points of x-intercepts would be the results. For these points, the x-coordinate shall serve the purpose of the results.

To cross check, put these values in the given equation to confirm that the left-hand side and right-hand side matches.

Calculation :

Using the graphing calculator, the graph for the given function is:

In this, the parabola x-intercepts at $\left(3,0\right)$ , right to the origin. This value of the x-intercept shall be the time for which the ball was in the air.

Now, cross check by putting these values of x-coordinate in the given equation.

  $-16t^2+47t+3=0$

This is the given equation.

  $=>-16{\left(3\right)}^2+47\left(3\right)+3=0$

Putting $t=3$

  $=>-144+141+3=0$

Simplifying this

  $=>0=0$

Thus the test is a success and the result is 3 seconds .

b.

To determine

To calculate the y-intercept and explanation of the same.

Answer to Problem 28PPS

The y-intercept for the $\left(0,3\right)$ . This signifies the height from which the ball was hit from a height of 3 feet.

Explanation of Solution

Given information :

The function provided for the path of the ball is $h=-16t^2+47t+3$ .

Formula used :

To calculate the y-intercept, replace the value of t with 0. The resultant value of h will be the y-intercept.

Calculation :

Replace the value of t in the given equation

  $h=-16t^2+47t+3$

This is the given equation.

  $=>-16{\left(0\right)}^2+47\left(0\right)+3=0$

Putting $t=0$

  $=3$

Thus, the y-intercept is $\left(0,3\right)$ . This signifies that the ball was hit at a level of 3 feet from the ground.