Chapter SH, Problem 9.14EP

To determine

To find: The time when the marble will hit the ground if the equation of the height is $h=-16t^2+42t+5.5$ .

Answer to Problem 9.14EP

The marble will hit the ground after 2.75 seconds.

Explanation of Solution

Given information:

The given equation of the height is $h=-16t^2+42t+5.5$ .

Calculation:

The given equation is a quadratic equation.

  $h=-16t^2+42t+5.5$

Write the standard form of the quadratic equation.

  $y=a{x}^2+bx+c$

Compare the given equation with the standard form of the quadratic equation.

  $a=-16,b=42,c=5.5$

Write the equation for the axis of the symmetry.

  $x=\frac{-b}{2a}$

Substitute $42$ for $b$ and $-16$ for $a$ .

  $\begin{array}{c}x=\frac{-\left(42\right)}{2\left(-16\right)}\\ =\frac{42}{32}\\ =1.313\end{array}$

So, the equation for the axis of the symmetry is $x=1.313$ .

The vertex is equal to $\left(x,y\right)$ . To calculate the value of $y$ , substitute $1.313$ for $x$ in the given equation.

  $\begin{array}{c}y=-16{\left(1.313\right)}^2+42\left(1.313\right)+5.5\\ =33.063\end{array}$

So, the vertex of the equation is $\left(1.313,33.063\right)$ .

The y -intercept of the graph of the equation is equal to the value of $c$ . So, the y -intercept of the graph of the equation is $5.5$ .

Draw the graph of the equation.

  

From the graph, the graph intersects the x -axis at $-0.125$ and $2.75$ . Neglect the negative value of time. So, the marble will hit the ground after 2.75 seconds.

Therefore, the marble will hit the ground after 2.75 seconds.