Chapter 6, Problem 84RE

Solution

To find:The time in seconds by which the ball hit the ground after throwing. Distance of ball from the initial position (from where the ball is thrown).

The time in seconds by which the ball hit the ground after throwing is $t\approx 0.711\sec$ .

Sharon will be at a distance of approximately $46.75\text{ft}$ from the point where ball hit the ground.

Given information:

Sharon throws the ball from the 4 ft above the ground with an initial velocity $66\text{ft}/\sec$ at and angle $5°$ with the horizontal.

Formula used:

The quadratic formula, $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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, $v_0=66\text{ft}/\sec$ , $\theta =5°$ , and $y_0=4\text{ft}$ .

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

  $x=\left(66\cos 5°\right)t$ and $y=-16t^2+\left(66\sin 5°\right)t+4$

Solve for the initial condition $y=0$ (means baseball is at ground).

  $0=-16t^2+\left(66\sin 5°\right)t+4$

Apply the quadratic formula $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ with $a=-16$ , $b=66\sin 5°$ , and $c=4$ .

  $\begin{array}{l}t=\frac{-\left(66\sin 5°\right)\pm \sqrt{{\left(66\sin 5°\right)}^2-4\left(-16\right)\left(4\right)}}{2\left(-16\right)}\\ t\approx -0.35,0.711\end{array}$

Since time cannot be negative, take $t\approx 0.711\sec$ .

Substitute $t\approx 0.711\sec$ into $x=\left(66\cos 5°\right)t$ and evaluate the value of $x$ .

  $\begin{array}{c}x=\left(66\cos 5°\right)\left(0.711\right)\\ x\approx 46.78\text{ft}\end{array}$