Chapter 10.7, Problem 50AYU

To determine

To find:

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

Answer to Problem 50AYU

a. $y=-16t²+40t+6$

Explanation of Solution

Given:

Alice throws a ball straight up with an initial speed of 40 feet per second from a height of 5 feet.

Formula used:

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

Calculation:

$v_o=40\text{ feet}/\text{sec}$, $h=5\text{ feet}$, $g=32\text{ feet}/\text{sec}$, $\theta =90°$

a. $x=40\text{cos}90*t=0$

$y=-\frac{1}{2}*32*t²+40*\text{ sin}90*t+5$

$y=-16t²+40t+6$

To determine

To find:

b. How long is the ball in the air?

Answer to Problem 50AYU

b. The ball is in the air for $2.62$ secs.

Explanation of Solution

Given:

Alice throws a ball straight up with an initial speed of 40 feet per second from a height of 5 feet.

Formula used:

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

Calculation:

$v_o=40\text{ feet}/\text{sec}$, $h=5\text{ feet}$, $g=32\text{ feet}/\text{sec}$, $\theta =90°$

b. When the ball is in the air $y=0$.

Solving the equation $y=-16t²+40t+5=0$,

$t=\left(-40\pm \sqrt{{40}^2}-4*-16*5\right)/2*-16$

$t=\left(-40\pm \surd 1920\right)/-32$

$t=-0.12$ or $2.62$

Hence, the ball is in the air for $2.62$ secs.

To determine

To find:

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

Answer to Problem 50AYU

c. Maximum height $=30$ feet.

Explanation of Solution

Given:

Alice throws a ball straight up with an initial speed of 40 feet per second from a height of 5 feet.

Formula used:

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

Calculation:

$v_o=40\text{ feet}/\text{sec}$, $h=5\text{ feet}$, $g=32\text{ feet}/\text{sec}$, $\theta =90°$

c. The maximum height occurs at the vertex.

$t=-b/2a=\frac{40}{2\left(-16\right)}=1.25$ seconds

Maximum height is $y=-16t²+50t+6$ when $t=1.5625$ sec.

$y=-16{\left(1.25\right)}^2+40*1.25+5=30$

Maximum height $=30$ feet.

To determine

To find:

d. Simulate the motion of the ball by graphing the equations found in part a.

Answer to Problem 50AYU

d.

Explanation of Solution

Given:

Alice throws a ball straight up with an initial speed of 40 feet per second from a height of 5 feet.

Formula used:

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

Calculation:

$v_o=40\text{ feet}/\text{sec}$, $h=5\text{ feet}$, $g=32\text{ feet}/\text{sec}$, $\theta =90°$

d. We have plotted the graph above with $x=3$ so that the line is not on the top of $y\text{-axis}$.