Chapter 10.7, Problem 56AYU

Projectile Motion Suppose that Karla hits a golf ball off a cliff 300 meters high with an initial speed of 40 meters per second at an angle of ${45}^{\circ }$ to the horizontal on the Moon (gravity on the Moon is one-sixth of that on Earth).

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

b. How long is the ball in the air?

c. Determine the horizontal distance that the ball travels.

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

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

To determine

To find:

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

Answer to Problem 56AYU

a. $x=20\surd 2t$, $y=-4.9/6t^2 + 20\sqrt{2}t + 300$

Explanation of Solution

Given:

Suppose that Karla 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 on the Moon (gravity on the Moon is one-sixth of that on Earth).

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

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

$h=300$, $g=9.8/6$

$y=-\frac{1}{2} \&*\#x00A0;\frac{9.8}{6}t² + (40\sin 45)t + 300$

$y=-4.9t² + 40t\text{sin}45 + 300$

$y=-\frac{4.9}{6}{t}^2 + 20\sqrt{2}t + 300$

To determine

To find:

b. How long is the ball in the air?

Answer to Problem 56AYU

b. $43.15$ secconds.

Explanation of Solution

Given:

Suppose that Karla 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 on the Moon (gravity on the Moon is one-sixth of that on Earth).

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:

$y=-\frac{4.9}{6}t² + 20\sqrt{2}t + 300$

$0=-\frac{4.9}{6}{t}^2 + 20\sqrt{2}t + 300$

$t=\frac{-20\surd 2 \pm \sqrt{{\left(20\surd 2\right)}^2 - 4\left(-4.9/6\right)\left(300\right)}}{2\left(-4.9\right)}$

$t=-8.51374$ or $43.15$

The ball is in the air for $43.15$ seconds.

To determine

To find:

c. The horizontal distance that the ball travels.

Answer to Problem 56AYU

c. $1220.5$ meters.

Explanation of Solution

Given:

Suppose that Karla 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 on the Moon (gravity on the Moon is one-sixth of that on Earth).

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 $x=\left(v_o\text{cos}\theta \right)t$

$x=20\surd 2 \&*\#x00A0;43.15 \approx 1220.5$ meters.

To determine

To find:

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

Answer to Problem 56AYU

d. $y \approx 544.9$ metres.

Explanation of Solution

Given:

Suppose that Karla 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 on the Moon (gravity on the Moon is one-sixth of that on Earth).

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.

$t=-\frac{b}{2a}=-\frac{20\surd 2}{2\left(-4.9/6\right)} \approx 17.32$ seconds.

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

$y=-4.9\left(17.32\right)² + 20\sqrt{2}\left(17.32\right) + 300$

$y \approx 544.9$ metres.

To determine

To find:

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

Answer to Problem 56AYU

Explanation of Solution

Given:

Suppose that Karla 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 on the Moon (gravity on the Moon is one-sixth of that on Earth).

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.