Chapter 6, Problem 87RE

Solution

a)

To find: The ball’s maximum height from the ground.

The maximum height $77.6\text{ft}$ .

Given information:

The initial velocity of the ball is $85\text{ft}/\text{sec}$ . The ball leaves the ground at 15 yd line at an angle of $56°$ .

Calculation:

When an object is in projectile motion at an angle of $\theta$ , the its horizontal and vertical velocity is given by $V_h=V\cos \theta$ and $V_v=V\sin \theta$ respectively.

Consequently, in the given case, horizontal velocity will be $85\cos 56°$ and vertical velocity will be $85\sin 56°$ .

The vertical position of an object affected by gravity is given by $s=-16t^2+v_{0}t+h_0$ , where $v_0$ is the initial velocity, $h_0$ is the initial time, and $t$ is time.

Consequently, in the given case the equation will be:

  $\begin{array}{l}s=-16t^2+\left(85\sin 56°\right)t+0\\ s=-16t^2+\left(85\sin 56°\right)t\end{array}$

The maximum height attains at a point where the axis of symmetry for a parabola is $t=-\frac{b}{2a}$ .

In this case, $b=85\sin 56°$ and $a=-16$ . So,

  $\begin{array}{c}t=-\frac{85\sin 56°}{2\left(-16\right)}\\ t\approx 2.2\sec \end{array}$

Substitute $t=2.2$ into the equation $s=-16t^2+\left(85\sin 56°\right)t$ and solve for $s$ to determine the maximum height.

  $\begin{array}{c}s=-16{\left(2.2\right)}^2+\left(85\sin 56°\right)\left(2.2\right)\\ s\approx 77.6\text{ft}\end{array}$

Therefore, the maximum height $77.6\text{ft}$ .

b)

To find:The “hang time” means the time during which the ball was in air.

The “hang time” is $4.4\sec$ .

Given information:

The initial velocity of the ball is $85\text{ft}/\text{sec}$ . The ball leaves the ground at 15 yd line at an angle of $56°$ .

Calculation:

Determine the zeros (other than $t=0$ ) in order to know that when the ball is on ground.

Substitute 0 for $s$ into $s=-16t^2+\left(85\sin 56°\right)t$ and solve for $t$ .

  $\begin{array}{c}0=-16t^2+\left(85\sin 56°\right)t\\ 0=t\left(-16t+85\sin 56°\right)\\ 0=-16t+85\sin 56°\\ 16t=85\sin 56°\\ t=\frac{85\sin 56°}{16}\\ t\approx 4.4\sec \end{array}$

Since, the initial height was taken to be 0 at $t=0$ , time from start to the maximum hight could be doubled (because the axis of symmetry is the middle point of the parabola).