Chapter 10.7, Problem 55AYU

To determine

To find:

a. Parametric equations that model the position of the ball as a function of time.

Answer to Problem 55AYU

a. $x=20\surd 2t$, $y=-4.9t^2 + 20\sqrt{2}t + 300$

Explanation of Solution

Given:

Suppose that Adam hits a golf ball off a cliff 300 meters high with an initial speed of 40 meters per second at an angle of $45°$ to the horizontal.

Formula used:

$x=\left(v_o\text{cos}\theta \right)t$, $y=-\frac{1}{2}gt² + \left(v_o\text{sin}\theta \right)t + h$

Calculation:

$v_o=40$, $\theta =45°$

a. $x=\left(40\text{cos}45\right)t$

$x=\left(40 \&*\#x00A0;\frac{1}{\sqrt{2}}\right)t=20\surd 2t$

$h=300$

$y=-\frac{1}{2} \&*\#x00A0;9.8t² + (40\sin 45)t + 300$

$y=-4.9t² + 40t\text{sin}45 + 300$

$y=-4.9t^2 + 20\sqrt{2}t + 300$

To determine

To find:

b. How long is the ball in the air?

Answer to Problem 55AYU

b. $11.23$ secconds.

Explanation of Solution

Given:

Suppose that Adam hits a golf ball off a cliff 300 meters high with an initial speed of 40 meters per second at an angle of $45°$ to the horizontal.

Formula used:

$x=\left(v_o\text{cos}\theta \right)t$, $y=-\frac{1}{2}gt² + \left(v_o\text{sin}\theta \right)t + h$

Calculation:

$v_o=40$, $\theta =45°$

b. The ball is in the air until $y=0$, Solve:

$y=-4.9t² + 20\sqrt{2}t + 300$

$0=-4.9t^2 + 20\sqrt{2}t + 300$

$t=\frac{-20\surd 2 \pm \sqrt{{\left(20\surd 2\right)}^2 - 4\left(-4.9\right)\left(300\right)}}{2\left(-4.9\right)}$

$t=-5.45$ or $11.23$

The ball is in the air for $11.23$ seconds.

To determine

To find:

c. The horizontal distance that the ball travels.

Answer to Problem 55AYU

c. $317.63$ meters.

Explanation of Solution

Given:

Suppose that Adam hits a golf ball off a cliff 300 meters high with an initial speed of 40 meters per second at an angle of $45°$ to the horizontal.

Formula used:

$x=\left(v_o\text{cos}\theta \right)t$, $y=-\frac{1}{2}gt² + \left(v_o\text{sin}\theta \right)t + h$

Calculation:

$v_o=40$, $\theta =45°$

c. Horizontal distance $x=\left(v_o\text{cos}\theta \right)t$

$x=20\surd 2 \&*\#x00A0;11.23 \approx 317.63$ meters.

To determine

To find:

d. When is the ball at its maximum height? Determine the maximum height of the ball.

Answer to Problem 55AYU

d. $y \approx 340.82$ metres.

Explanation of Solution

Given:

Suppose that Adam hits a golf ball off a cliff 300 meters high with an initial speed of 40 meters per second at an angle of $45°$ to the horizontal.

Formula used:

$x=\left(v_o\text{cos}\theta \right)t$, $y=-\frac{1}{2}gt² + \left(v_o\text{sin}\theta \right)t + h$

Calculation:

$v_o=40$, $\theta =45°$

d. The maximum height is at vertex of the quadratic function.

$t=-\frac{b}{2a}=- \frac{20\surd 2}{2\left(-4.9\right)} \approx 2.89$ seconds.

Substituting the value in $y=-4.9t² + 20\sqrt{2}t + 300$

$y=-4.9\left(2.89\right)² + 20\sqrt{2}\left(2.89\right) + 300$

$y \approx 340.82$ metres.

To determine

To find:

e. Using a graphing utility, simultaneously graph the equations found in part (a).

Answer to Problem 55AYU

Explanation of Solution

Given:

Suppose that Adam hits a golf ball off a cliff 300 meters high with an initial speed of 40 meters per second at an angle of $45°$ to the horizontal.

Formula used:

$x=\left(v_o\text{cos}\theta \right)t$, $y=-\frac{1}{2}gt² + \left(v_o\text{sin}\theta \right)t + h$

Calculation:

$v_o=40$, $\theta =45°$

e. Graph of the equation found in a.