Chapter 6, Problem 88RE

Solution

To find:Initial velocity of the ball to cross the fence.

The initial velocity of the ball is approximately equal to $125\text{ft}/\sec$ .

Given information:

The height of fence is15 ft, which is 400 ft away from home plate. Brian hits the ball from $2.5\text{ft}$ above the ground at an angle of $30°$ from the ground.

Formula used:

The trigonometric ratios $\cos 30°=\frac{\sqrt{3}}{2}$ and $\sin 30°=\frac{1}{2}$ .

Calculation:

It is known that the path of an object in projectile motion is modelled by the parametric equations $x=\left(v_0\cos \theta \right)t$ and $y=-16t^2+\left(v_0\sin \theta \right)t+y_0$ , where $y_0$ is the point above the ground from where the object is thrown, $v_0$ is the initial speed in $\text{ft}/\sec$ at an angle $\theta$ with horizontal.

In the given situation, $x=400\text{ft}$ , $\theta =30°$ , $y=15\text{ft}$ and $y_0=2.5\text{ft}$ .

Substitute these values into the parametric equations $x=\left(v_0\cos \theta \right)t$ and $y=-16t^2+\left(v_0\sin \theta \right)t+y_0$ .

  $400=\left(v_0\cos 30°\right)t$ and $15=-16t^2+\left(v_0\sin 30°\right)t+2.5$

Simplify the equation $400=\left(v_0\cos 30°\right)t$ for $t$ .

  $\begin{array}{c}400=\left(v_0\cos 30°\right)t\\ t=\frac{400}{v_0\cos 30°}\end{array}$

Substitute $t=\frac{400}{v_0\cos 30°}$ into the equation $15=-16t^2+\left(v_0\sin 30°\right)t+2.5$ and solve it for $v_0$ . Use $\cos 30°=\frac{\sqrt{3}}{2}$ and $\sin 30°=\frac{1}{2}$ .

  $\begin{array}{c}15=-16{\left(\frac{400}{v_0\cos 30°}\right)}^2+\left(v_0\sin 30°\right)\left(\frac{400}{v_0\cos 30°}\right)+2.5\\ 15=-16{\left(\frac{400}{v_0\left(\frac{\sqrt{3}}{2}\right)}\right)}^2+\left(\frac{1}{2}{v}_0\right)\left(\frac{400}{v_0\left(\frac{\sqrt{3}}{2}\right)}\right)+2.5\\ 15=-16\left(\frac{160,000\cdot 4}{3v_0{}^2}\right)+\frac{400}{\sqrt{3}}+2.5\\ 0=-\frac{10,240,000}{3v_0{}^2}+\frac{400}{\sqrt{3}}+2.5-15\\ \end{array}$

Further, simplify.

  $\begin{array}{c}\frac{10,240,000}{3v_0{}^2}=\frac{400}{\sqrt{3}}-12.5\\ v_0{}^2=\frac{10,240,000}{3\left(\frac{400}{\sqrt{3}}-12.5\right)}\\ v_0=\sqrt{\frac{10,240,000}{3\left(\frac{400}{\sqrt{3}}-12.5\right)}}\\ v_0\approx 125\text{ft}/\sec \end{array}$

Therefore, the initial velocity of the ball is approximately equal to $125\text{ft}/\sec$ .