Chapter 1, Problem 41REAE

Solution

To determine the time for which the ball remained in the air.

The ball remained in the air for $\frac{\sqrt{34}-5}{4}$ seconds.

Given:

Given the initial vertical velocity of the ball:

  $v_0=-40\text{ft/s}$ .

Given the initial height of the ball:

  $h_0=9\text{ft}$ .

Concept Used:

Equation modelling vertical motion:

The equation modelling the height $h\left(t\right)$ of a vertically moving object, which had an initial velocity of $v_0\text{ft/s}$ and initial height of $h_0\text{ft}$ , after $t$ seconds:

  $h\left(t\right)=-16t^2+v_{0}t+h_0$ .

Quadratic formula:

Given a quadratic equation $a{t}^2+bt+c=0$ .

Then, the solutions of the equation:

  $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Calculation:

Substitute $v_0=-40$ and $h_0=9$ in the equation $h\left(t\right)=-16t^2+v_{0}t+h_0$ modelling the vertical motion:

  $\begin{array}{l}h\left(t\right)=-16t^2+\left(-40\right)t+\left(9\right)\\ h\left(t\right)=-16t^2-40t+9\end{array}$ .

It is asked to determine the time for which the ball remained in air.

It is the value of $t$ for which the height $h\left(t\right)$ is nullified.

That is, it is the value of $t$ for which $h\left(t\right)=0$ .

Substitute $h\left(t\right)=0$ in the equation $h\left(t\right)=-16t^2-40t+9$ :

  $0=-16t^2-40t+9$ .

Solve the equation $0=-16t^2-40t+9$ for $t$ :

  $\begin{array}{l}t=\frac{-\left(-40\right)\pm \sqrt{{\left(-40\right)}^2-4\left(-16\right)\left(9\right)}}{2\left(-16\right)}\\ t=\frac{40\pm \sqrt{2176}}{-32}\\ t=\frac{40\pm 8\sqrt{34}}{-32}\\ t=-\frac{5+\sqrt{34}}{4},\frac{\sqrt{34}-5}{4}\end{array}$ .

Now, $t$ represents the time.

Hence, $t$ must be positive.

Then, it follows that, $t=\frac{\sqrt{34}-5}{4}$ .

Thus, the ball remained in the air for $\frac{\sqrt{34}-5}{4}$ seconds.

Conclusion:

The ball remained in the air for $\frac{\sqrt{34}-5}{4}$ seconds.