Chapter 9, Problem 8MCQ

a.

To determine

To calculate the height(h) of the ball at one second i.e. $t=1$

Answer to Problem 8MCQ

The height reached by the ball is $h=74$ meters.

Explanation of Solution

Given information :

The function provided is $h=-16t^2+90t$

Formula used :

The height of the football after one second i.e. at $t=1$ , can be calculated by putting the value of t in the quadratic equation $h=-16t^2+90t$ which gives the height h of the ball after t seconds.

Calculation :

Quadratic equation given:

  $h=-16t^2+90t$

Putting value of t as 1.

  $h=-16{(1)}^2+90(1)$

Solving this.

  $h=74$

Thus the height of the ball after one second is $h=74$ meters.

b.

To determine

To calculate the time when the ball is at the maximum height.

Answer to Problem 8MCQ

The ball is at the maximum height at $t=2.813$ .

Explanation of Solution

Given information :

The function provided is $h=-16t^2+90t$

Formula used :

The graph of a quadratic equation ( $y=a{x}^2+bx+c$ ) is a parabola. The parabola is symmetric in nature and the point of symmetry is known as the vertex. This vertex lies on the line of symmetry and is symmetric about this line.

This vertex point shall be:

1. Highest point (if $a<0$) and be called the maximum. Here, the graph opens downwards.
2. Lowest point (if $a>0$) and be called the minimum. Here, the graph opens upwards.

Here, a is less than zero, hence the parabola opens downwards, and has a maximum value.

A parabola always points to infinity, either negative or positive.

To graph a quadratic equation using table of values, compute the ‘y’ values for the corresponding ‘x’ values. For this, substitute the values of ‘x’ with negative as well as positive integers and check for the vertex.

Axis of symmetry for the given function is

  $x=-\frac{b}{2\times a}$

Calculation :

Substituting the values from the given equation.

  $t=-\frac{(90)}{2\times (-16)}$

Simplifying.

  $t=\frac{45}{16}$

This corresponds to the x -coordinate of the vertex

To graph a quadratic equation using table of values, compute the ‘y’ values for the corresponding ‘x’ values. For this, substitute the values of ‘x’ with negative as well as positive integers and check for the vertex.

For example, to compute the value of ‘h’ when $t=\frac{45}{16}$ , substitute the value of ‘t’ in equation $h=-16t^2+90t$

On substituting the value of the ‘t’ as $t=\frac{45}{16}$ , it is,

  $h=-{(\frac{45}{16})}^2+90(\frac{45}{16})$

Finally, the value of ‘h’ is

  $=126.56$

Therefore, the function $h=-16t^2+90t$ has a maximum value of h $=126.56$ after $t=2.813$ seconds

c.

To determine

To calculate the time when the height of the ball is 0 feet.

Answer to Problem 8MCQ

The ball is 0 feet high at $t=0\text{ and }5.625$ seconds.

Explanation of Solution

Given information :

The function provided is $h=-16t^2+90t$ .

Formula used :

To calculate the time when height of ball is 0 feet, use the formula to find roots of a quadratic equation. This can be found by using $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . Here ‘a’, ‘b’ and ‘c’ are the co-efficient of the given equation.

Here, $a=-16,b=90\text{ and}c=0$ .

Calculation :

  $t=\frac{-90\pm \sqrt{{90}^2-4\times (-16)\times 0}}{2(-16)}$

Putting the values in the formula.

  $t=\frac{-90\pm \sqrt{{90}^2}}{-32}$

Removing square root.

  $t=\frac{-90\pm 90}{-32}$

One of the roots is

  $t=0$

This is the first root.

Second root is.

  $t=\frac{-90-90}{-32}$

Dividing the fraction.

  $t=5.625$

Thus, the ball is 0 feet high at $t=0\text{ and }5.625$ seconds.

This implies the ball is at 0 feet when it is kicked and when it hits the ground 5.625 seconds later

d.

To determine

To state the mathematical practice used to solve the problem.

Explanation of Solution

Given information :

The solutions of a quadratic equation are called roots.

If the quadratic function $y=a{x}^2+bx+c$   has real solutions can be found out by using the quadratic formula:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The formula $b^2-4ac$ , also known as the discriminant can be used to determine the number of roots of solutions in a quadratic equation. There are three cases:

• $b^2-4ac<0$: The equation has 0 real solutions. The graph does not cross the  x -axis.
• $b^2-4ac=0$: The equation has 1 real solution. The graph touches the  x -axis at one point.
• $b^2-4ac>0$: The equation has 2 real solutions. The graph crosses through the  x -axis at two points.

A real number  x  will be called a solution or a root iff it satisfies the equation, meaning $a{x}^2+bx+c=0$ .