Q8. - Algebra 1

Question

A soccer ball is kicked from ground level with an initial upward velocity of 90 feet per second. The equation $h=-16t^2+90t$ gives the height h of the ball after t seconds.

 

a. What is the height of the ball after one second?

b. How many seconds will it take for the ball to reach its maximum height?

c. When is the height of the ball 0 feet? What do these points represents in this situation?

Step-by-Step Solution

Answer

a. The height of the ball after one second is 74 feet.

b. The ball will take 2.8125 seconds to reach its maximum height.

c. After $t=0$ seconds and $t=5.625$ seconds the height of the ball is 0 feet.

In this situation these two points represents that the at the start $t=0$ seconds the ball is at the ground level and after $t=5.625$ seconds the ball will hit the ground level after travelling in the air.

Part a. Step 1. Define an equation.

An equation is a mathematical expression that contains an equal symbol $"="$.

Part a. Step 2. 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.

Part a. Step 3. Calculate the height of the ball after one second.

Substitute $t=1$ in $h=-16t^2+90t$.

$$\begin{gathered} h=-16{\left(1\right)}^2+90\left(1\right) \\ h=-16\left(1\right)+90 \\ h=-16+90 \\ h=74 \end{gathered}$$

 

Hence the height of the ball after one second is 74 feet.

Part b. Step 1. Define an equation.

An equation is a mathematical expression that contains an equal symbol $"="$.

Part b. Step 2. 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.

Part b. Step 3. 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 $a<0$.

Part b. Step 4. Calculate the time taken by the ball to reach its maximum height.

Observe the function $h=-16t^2+90t$.

$a=-16, b=90$ and $c=0$

Since, $a<0$

So, the graph of this function opens downward and has a maximum value at $t=-\frac{b}{2a}$.

Substitute $a=-16$ and $b=90$ in $t=-\frac{b}{2a}$.

$$\begin{gathered} t=-\left(\frac{90}{2\left(-16\right)}\right) \\ t=-\left(\frac{90}{-32}\right) \\ t=\frac{90}{32} \\ t=2.8125 \end{gathered}$$

 

Hence, the ball will take 2.8125 seconds to reach its maximum height. 

Part c. Step 1. Define an equation.

An equation is a mathematical expression that contains an equal symbol $"="$.

Part c. Step 2. 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.

Part c. Step 3. Calculate the time at which the height of the ball is 0 feet.

If the height of the ball is 0 feet, then $h=0$.

Substitute $h=0$ in $h=-16t^2+90t$.

width="190" height="116" style="max-width: none; vertical-align: -59px;" $$\begin{gathered} 0=-16t^2+90t \\ -2t\left(8t-45\right)=0 \\ t\left(8t-45\right)=\frac{0}{-2} \\ t\left(8t-45\right)=0 \end{gathered}$$

$t=0$ or $8t=45$

$t=0$ or $t=\frac{45}{8}$

$t=0$ or $t=5.625$

$t=0,5.625$

 

Hence, after $t=0$ seconds and $t=5.625$ seconds the height of the ball is 0 feet.

In this situation these two points represents that the at the start $t=0$ seconds the ball is at the ground level and after $t=5.625$ seconds the ball will hit the ground level after travelling in the air.