Chapter 10.7, Problem 50AYU

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

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

b. How long is the ball in the air?

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

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

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 $y\text{-axis}$.