Q63. - Algebra 1

Question

The average amateur golfer can hit a ball when an initial velocity of  meters per second. If the ball is hit straight up, the height can be modeled by the equation $h=-4.9t^2+31.3t,$ where $h$ is the height of the ball, meters, after $t$ seconds.

 

1. Graph this equation.
2. At what height is the ball hit?
3. What is the maximum height of the ball?
4. How long did it take for the ball to hit the ground?
5. e. State a reasonable range and domain for this situation.

Step-by-Step Solution

Answer

a. The graph of the function $h=-4.9t^2+31.3t$ is given below.

b. The height at which the ball hit is $0 \mathrm{feet}$

c. The maximum height of the ball is $50 \mathrm{meter}$

d. For the given situation, the domain is $\left\{0\le t\le 6.4\right\}$ and the range is $\left\{0\le h\le 50\right\}$.

e. The time taken by the ball to hit the ground is $6.4 \mathrm{seconds}$

Step 1. Define the standard form of the quadratic function.

A quadratic function, which is written in the form, $y=a{x}^2+bx+c$, where, $a\ne 0$ is called the standard form of the quadratic function.

Step 2. Define the maximum or minimum point of the function y = a x 2 + b x + c .

The graph of the function $y=a{x}^2+bx+c$

Opens upward and has a minimum value at $x=-\frac{b}{2a}$, when $a>0$.

Opens downward and has a maximum value at $x=-\frac{b}{2a}$,  when $a<0$.       

Step 3. Calculation.

a.

The $y$-intercept of the function $y=a{x}^2+bx+c$ is always at $c$.

Compare the quadratic function $h=-4.9t^2+31.3t$ with the standard quadratic function $h=a{t}^2+bt+c$.

$a=-4.9,b=31.3,c=0$

Substitute $a=-4.9$ and $b=31.3$ in $t=-\frac{b}{2a}$.

$$\begin{gathered} t=-\left(\frac{31.3}{2\left(-4.9\right)}\right) \\ t=-\left(-\frac{31.3}{9.8}\right) \\ t\approx 3.2 \end{gathered}$$

Since $a<0$.

So, the graph opens downward and has a maximum value $t=3.2$.

Since the $y$-intercept is given by $c$.

So, the $y$-intercept is 0.

The graph of the function $h=-4.9t^2+31.3t$ is shown below.

b.

The ball will hit where the time is 0.

Substitute $t=0$ in $h=-4.9t^2+31.3t$.

$$\begin{gathered} h=-4.9{\left(0\right)}^2+31.3\left(0\right) \\ h=0+0 \\ h=0 \end{gathered}$$

Hence, the height at which the ball hit is 0 feet.

 

c.

Compare the quadratic function $h=-4.9t^2+31.3t$ with the standard quadratic function $h=a{t}^2+bt+c$.

$a=-4.9,b=31.3,c=0$

Substitute $a=-4.9$ and $b=31.3$ in $t=-\frac{b}{2a}$.

$$\begin{gathered} t=-\left(\frac{31.3}{2\left(-4.9\right)}\right) \\ t=-\left(-\frac{31.3}{9.8}\right) \\ t\approx 3.2 \end{gathered}$$

Since $a<0$.

So, the graph opens downward and has a maximum value $t=3.2$.

Substitute $t=3.2$ in $h=-4.9t^2+31.3t$.

$$\begin{gathered} h=-4.9{\left(3.2\right)}^2+31.3\left(3.2\right) \\ h=-4.9\left(10.24\right)+100.16 \\ h=-50.176+100.16 \\ h=49.984 \\ h\approx 50 \end{gathered}$$

Hence, the maximum height of the ball is $50 \mathrm{meters}$.

       

d.

Compare the quadratic function $h=-4.9t^2+31.3t$ with the standard quadratic function $h=a{t}^2+bt+c$.

$a=-4.9,b=31.3,c=0$

Substitute $a=-4.9$ and $b=31.3$ in $t=-\frac{b}{2a}$.

$$\begin{gathered} t=-\left(\frac{31.3}{2\left(-4.9\right)}\right) \\ t=-\left(-\frac{31.3}{9.8}\right) \\ t\approx 3.2 \end{gathered}$$

Since $a<0$.

So, the graph opens downward and has a maximum value $t=3.2$.

Since the time taken by the ball to hit the ground will be twice the time taken by the ball to reach the maximum point.

So, the time is taken by the ball to hit the ground $=2\times 3.2=6.4$ seconds

Hence, the time taken by the ball to hit the ground is $6.4 \mathrm{seconds}$.

e.

The domain is the set of all of the possible values of the independent variable $x$.

The range is the set of all the possible values of the dependent variable $y$.

 

Since, when the ball is hit straight up, the time, $t=0$ and when the ball reaches the ground, the time, $t=6.4$.

So, the domain $=\left\{0\le t\le 6.4\right\}$

Since the ball is initially at $0 \mathrm{meter}$ height and reaches a maximum height of $50 \mathrm{meter}$.

So, the range $=\left\{0\le h\le 50\right\}$

Hence, for the given situation, the domain is $\left\{0\le t\le 6.4\right\}$ and the range is $\left\{0\le h\le 50\right\}$.