Chapter 10.7, Problem 49AYU

To determine

To find:

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

Answer to Problem 49AYU

a. $y=-16t²+50t+6$

Explanation of Solution

Given:

Bob throws a ball straight up with an initial speed of 50 feet per second from a height of 6 feet.

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:

$v_o=50\text{ feet}/\text{sec}$, $h=6\text{ feet}$, $g=32\text{ feet}/\text{sec}$, $\theta =90°$

a. $x=50\text{cos}90*t=0$

$y=-\frac{1}{2}*32*t²+50*\text{ sin}90*t+6$

$y=-16t²+50t+6$

To determine

To find:

b. How long is the ball in the air?

Answer to Problem 49AYU

b. The ball is in the air for $3.24$ secs.

Explanation of Solution

Given:

Bob throws a ball straight up with an initial speed of 50 feet per second from a height of 6 feet.

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:

$v_o=50\text{ feet}/\text{sec}$, $h=6\text{ feet}$, $g=32\text{ feet}/\text{sec}$, $\theta =90°$

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

Solving the equation $y=-16t²+50t+6=0$,

$t=\left(-50\pm \sqrt{({50}^2}-4*-16*6)\right)/2*-16$

$t=\left(-50\pm \sqrt{2884}\right)/-32$

$t=-0.12$ or $3.24$

Hence, the ball is in the air for $3.24$ secs.

To determine

To find:

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

Answer to Problem 49AYU

c. Maximum height $=45.0625$ feet.

Explanation of Solution

Given:

Bob throws a ball straight up with an initial speed of 50 feet per second from a height of 6 feet.

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:

$v_o=50\text{ feet}/\text{sec}$, $h=6\text{ feet}$, $g=32\text{ feet}/\text{sec}$, $\theta =90°$

c. The maximum height occurs at the vertex.

$t=-b/2a=\frac{50}{2\left(-16\right)}=1.5625$ seconds

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

$y=-16{\left(1.5625\right)}^2+50*1.5625+6=45.0625$

Maximum height $=45.0625$ feet.

To determine

To find:

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

Answer to Problem 49AYU

d.

Explanation of Solution

Given:

Bob throws a ball straight up with an initial speed of 50 feet per second from a height of 6 feet.

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:

$v_o=50\text{ feet}/\text{sec}$, $h=6\text{ feet}$, $g=32\text{ feet}/\text{sec}$, $\theta =90°$

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