Chapter 9.1, Problem 65PPS

a.

To determine

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

Answer to Problem 65PPS

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 126 feet high.

Answer to Problem 65PPS

The time when the ball is 126 feet high is

Explanation of Solution

Given information :

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

Formula used :

To calculate the time when $h=126$ , 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, $a=16,b=-90\text{ and}c=126$

Calculation :

The given quadratic equation.

  $h=-16t^2+90t$

Putting the value of $h=126$ in the equation,

  $126=-16t^2+90t$

Simplifying the expression.

  $16t^2-90t+126=0$

Putting the values in the formula.

  $t=\frac{-(-90)\pm \sqrt{{(-90)}^2-4\times 16\times 126}}{2\times 16}$

Simplifying the expression.

  $t=\frac{90\pm \sqrt{{8100}^{}-8064}}{32}$

Further simplification

  $t=\frac{90\pm \sqrt{36}}{32}$

Removing the square root

  $t=\frac{90\pm 6}{32}$

First square root is

  $t=\frac{96}{32}$

Second square root is

  $t=\frac{84}{32}$

Therefore in decimal form $t=3\text{ and }2.625$

Thus, the ball is 126 feet high at $t=2.625\text{ and 3}$ seconds.

c.

To determine

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

Answer to Problem 65PPS

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