Chapter 6, Problem 85RE

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 $t\approx 1.06\sec$ .

Diego will be at a distance of approximately $68.94\text{ft}$ from the point where ball hit the ground.

Given information:

Diego throws the ball from the $3.5\text{ft}$ above the ground with an initial velocity $66\text{ft}/\sec$ at an angle $12°$ 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, $v_0=66\text{ft}/\sec$ , $\theta =12°$ , and $y_0=3.5\text{ft}$ .

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

  $x=\left(66\cos 12°\right)t$ and $y=-16t^2+\left(66\sin 12°\right)t+3.5$

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

  $0=-16t^2+\left(66\sin 12°\right)t+3.5$

Apply the quadratic formula $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ with $a=-16$ , $b=66\sin 12°$ , and $c=3.5$ .

  $\begin{array}{l}t=\frac{-\left(66\sin 12°\right)\pm \sqrt{{\left(66\sin 12°\right)}^2-4\left(-16\right)\left(3.5\right)}}{2\left(-16\right)}\\ t\approx -0.2057,1.0634\end{array}$

Since time cannot be negative, take $t\approx 1.06\sec$ .

Substitute $t\approx 1.0634\sec$ into $x=\left(66\cos 12°\right)t$ and evaluate the value of $x$ .

  $\begin{array}{c}x=\left(66\cos 12°\right)\left(1.0634\right)\\ x\approx 68.94\text{ft}\end{array}$