Chapter 2, Problem 7CA

i.

To determine

To write: An equation for the path of the second ball.

Answer to Problem 7CA

The equation for the path of the second ball is,

  $y=-5{\left(x-3\right)}^2+51.5$

Here, x is time in seconds and y is height in feet.

Explanation of Solution

Given: The path for first ball is represented in the graph. The second ball is released 1.5 feet higher than the first ball and after 3 seconds reach its maximum height 5 feet lower than the first ball.

Concept Used:

The equation of the parabola with vertex $\left(h,k\right)$ is,

  $y=a{\left(x-h\right)}^2+k$ …… (1)

Calculation:

The height of the second ball is 5 feet lower than the first ball and at same time 3 seconds.

The vertex of the second ball is,

  $\left(3,56.5-5\right)=\left(3,51.5\right)$

The second ball is thrown from 1.5 feet above from where the first ball was thrown. The point from where the second ball was thrown is,

  $\left(0,5+1.5\right)=\left(0,6.5\right)$

Substitute the values of vertex $\left(h,k\right)$ and the point $\left(0,6.5\right)$ in equation (1) to obtain the value of a,

  $\begin{array}{c}y=a{\left(x-h\right)}^2+k\\ 6.5=a{\left(0-3\right)}^2+51.5\\ 6.5=9a+51.5\\ -51.5+6.5=9a\\ -45=9a\\ a=\frac{-45}{9}\\ =-5\end{array}$

Substitute the values $\left(h,k\right)=\left(3,51.5\right)\text{ and }a=-5$ to obtain the equation,

  $\begin{array}{c}y=a{\left(x-h\right)}^2+k\\ y=-5{\left(x-3\right)}^2+51.5\end{array}$

Here, x is time in seconds and y is height in feet.

(ii)

To determine

If the balls hit the ground together, how long were the balls in the air, and if the balls do not hit the ground together, which ball lands first.

Answer to Problem 7CA

First Ball reaches the ground first.

Explanation of Solution

Given: From the graph it is evident that first ball lands at the ground at 6 seconds

Concept Used:

When the ball reaches ground the height is equal to 0.

Calculation:

Put height equal to 0 in $y=-5{\left(x-3\right)}^2+51.5$ and find the time,

  $\begin{array}{c}y=-5{\left(x-3\right)}^2+51.5\\ 0=-5{\left(x-3\right)}^2+51.5\\ -51.5=-5{\left(x-3\right)}^2\\ 10.3={\left(x-3\right)}^2\\ x-3=\sqrt{10.3}\\ x=\sqrt{10.3}+3\\ =3.21+3\\ =6.21\end{array}$

Therefore, second ball reaches the ground at 6.21 seconds.