Chapter 9.1, Problem 63PPS

(a)

To determine

To Graph:

The equation.

Explanation of Solution

Given Information:

The average amateur golfer can hit a ball with an initial velocity of 31.3 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 , in meters , after t seconds.

Concept Used :

• Quadratic Functions:
• Quadratic functions are non-linear and are of the form:

  $f(x)=a{x}^2+bx+c;a\ne 0$

This is the standard form of Quadratic function.

• Shape of the graph of a Quadratic function is called Parabola.
• Parabolas are symmetric about a central line called Axis Of Symmetry.
• The axis of symmetry intersects a parabola only at one point , called the Vertex.
• Parent Function:

  $y=x^2$

• Standard Form :

  $f(x)=a{x}^2+bx+c;a\ne 0$

• Type of Graph:

Parabola

• Axis of Symmetry: Passes through the vertex and divides the parabola into two congruent halves.

  $x=-\frac{b}{2a}$

• y-intercept = The y coordinate of the point at which the graph intersects y-axis = c
• Graph:

  

• If a > 0 ,
• The graph of $a{x}^2+bx+c$ opens upward.
• The lowest point on the graph is the minimum.
• If a < 0 ,
• The graph of $a{x}^2+bx+c$ opens downward.
• The highest point on the graph is the maximum.
• The maximum or the minimum is the vertex.

Calculation:

• In the equation $h=-4.9t^2+31.3t$ ; comparing it with standard form of quadratic equation , we have a = -4.9 , b = 31.3 , c = 0 .
• Find equation of Axis of Symmetry :

  $\begin{array}{l}h=a{t}^2+bt+c\mathrm{...........................}[s\tan dard\_form]\\ t=-\frac{b}{2a}\mathrm{.....................................}[formula\_of\_axis\_of\_symmetry]\\ t=-\frac{31.3}{2(-4.9)}\approx \mathrm{3.2..........................}[a=-4.9,b=31.3]\end{array}$

The equation of axis of symmetry is $t=3.2$

• Find the Vertex and determine whether it is maximum or minimum:

Since axis of symmetry passes through the vertex , we have t coordinate of the vertex (t,h) is $t=3.2$

Substituting the value of $t=3.2$in the equation $h=-4.9t^2+31.3t$

  $\begin{array}{l}h=-4.9t^2+31.3t\\ h=-4.9{(}^{3.2}+31.3(3.2)\mathrm{...............}[t=3.2]\\ h=-50.2+\mathrm{100.2.........................}[simplify]\\ h\approx 50\end{array}$

The vertex is at $(3.2,50)$ .

Since the a = -4.9 < 0 .So, the graph opens downward .Hence, the vertex is the maximum.

• Find h-intercept :

The h-coordinate of the point at which the graph cuts the h-axis is the h-intercept.

h-intercept always occurs at (0,c) and c = 0 here.

So, h-intercept = 0 and the point is (0,0)

• The axis of symmetry cuts the parabola into two equal parts . So, if there is a point on one side, there is a corresponding point on the other side that is at a same distance from the axis of symmetry and the same h-value.

t	h
0	0
3.2	50
6.4	0

• Connect the points with a smooth curve.

Graph :

  

(b)

To determine

To Find:

Height at which the ball hit.

Answer to Problem 63PPS

0 feet

Explanation of Solution

Given Information:

The average amateur golfer can hit a ball with an initial velocity of 31.3 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 , in meters , after t seconds.

From part (a) , the vertex is (3.2,50) and it is the maximum value of the graph.

Calculation:

• In the equation $h=-4.9t^2+31.3t$; comparing it with standard form of quadratic equation , we have a = -4.9 , b = 31.3 , c = 0 .
• The ball is hit when the time is zero .
• Substituting t = 0 in the given equation:

  $\begin{array}{l}h=-4.9t^2+31.3t\mathrm{............................}[given\_equation]\\ h=-4.9(0)+31.3(0)\mathrm{......................}[t=0]\\ h=0\end{array}$

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

(c)

To determine

To Find:

The maximum height of the ball.

Answer to Problem 63PPS

50 feet

Explanation of Solution

Given Information:

The average amateur golfer can hit a ball with an initial velocity of 31.3 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 , in meters , after t seconds.

From part (a) , the vertex is (3.2,50) and it is the maximum value of the graph.

Concept Used :

• Quadratic Functions:
• Quadratic functions are non-linear and are of the form:

  $f(x)=a{x}^2+bx+c;a\ne 0$

This is the standard form of Quadratic function.

• y-intercept = The y coordinate of the point at which the graph intersects y-axis = c
• If a > 0 ,
• The graph of $a{x}^2+bx+c$ opens upward.
• The lowest point on the graph is the minimum.
• If a < 0 ,
• The graph of $a{x}^2+bx+c$ opens downward.
• The highest point on the graph is the maximum.
• The maximum or the minimum is the vertex.

Calculation:

• In the equation $h=-4.9t^2+31.3t$ ; comparing it with standard form of quadratic equation , we have a = -4.9 , b = 31.3 , c = 0 .
• To find the maximum height at which the ball reach:

Since, a = -4.9 < 0 . So, the graph of the function opens downward and the vertex is the maximum value.

The y-coordinate of the vertex is the maximum height the ball reach.

Vertex is (3.2,50) . So, the maximum height the ball reach is 50 feet.

(d)

To determine

To Find:

Time taken by the ball to hit the ground.

Answer to Problem 63PPS

6.4 seconds

Explanation of Solution

Given Information:

From part (a) , we have the vertex is $(3.2,50)$ and it is the maximum of the function.

Concept Used :

• Quadratic Functions:
• Quadratic functions are non-linear and are of the form:

  $f(x)=a{x}^2+bx+c;a\ne 0$

This is the standard form of Quadratic function.

• Shape of the graph of a Quadratic function is called Parabola.
• Parabolas are symmetric about a central line called Axis Of Symmetry.
• The axis of symmetry intersects a parabola only at one point , called the Vertex.
• Axis of Symmetry: Passes through the vertex and divides the parabola into two congruent halves.

Calculation:

• The vertex is (3.2,50)
• It takes the ball 3.2 seconds to reach the vertex, and due to symmetry it will take the ball another 3.2 seconds to hit the ground.
• So, it will take the ball 3.2 +3. 2 = 6.4 seconds to hit the ground.

(e)

To determine

To Find:

Domain and range of the function.

Answer to Problem 63PPS

Domain is

  $D=\{0\le t\le 6.4\}$

Range is

  $R=\{h:0\le h\le 50\}$

Explanation of Solution

Given Information:

From part (a) , we have the vertex is $(3.2,50)$ and it is the maximum of the function.

Concept Used :

• Quadratic Functions:
• Quadratic functions are non-linear and are of the form:

  $f(x)=a{x}^2+bx+c;a\ne 0$

This is the standard form of Quadratic function.

• If a > 0 ,
• The graph of $a{x}^2+bx+c$ opens upward.
• The lowest point on the graph is the minimum.
• If a < 0 ,
• The graph of $a{x}^2+bx+c$ opens downward.
• The highest point on the graph is the maximum.
• The maximum or the minimum is the vertex.
• The range is all real numbers less than or equal to the maximum value or all real numbers greater than or equal to the minimum value.
• The domain is all real numbers.

Calculation:

• When the ball is hit , time = 0 and when the ball reaches the ground , time = 6.4 .
So, the domain is $D=\{0\le t\le 6.4\}$ .
•  The ball is initially at 0 meter height , and reaches maximum height of 50 meters.

So, the range is $R=\{h:0\le h\le 50\}$