Chapter 10.7, Problem 49AYU

Projectile Motion Bob throws a ball straight up with an initial speed of 50 feet per second from a height of 6 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. 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 $y\text{-axis}$.