Chapter 9, Problem 67SGR

(a).

To determine

To find: The after how many seconds are the balls at the same height.

Answer to Problem 67SGR

The two balls will be at the same height after $1.5$ seconds.

Explanation of Solution

Given information:

Height of one ball $y=-16t^2+80t+25$

Height of one ball $y=-16t^2+30t+100$

Calculation:

Consider two balls that are launched into the air simultaneously from different altitudes. Their height y (feet) after t seconds is given by following equations.

  $y=-16t^2+80t+25\mathrm{..........}(1)$

And

  $y=-16t^2+30t+100\mathrm{..........}(2)$

Compute the time after which the balls will be at the same height, solve above two equations Substitute $y=-16t^2+80t+25$ from equation (1) in equation (2) $y=-16t^2+30t+100$ and solve for t,

  $\begin{array}{l}-16t^2+80t+25=-16t^2+30t+100\\ 50t=5\\ t=1.5\end{array}$

Therefore, the two balls will be at the same height after $1.5$ seconds.

(b).

To determine

To find: The height.

Answer to Problem 67SGR

The equal height of two balls is equal to $109$ feet.

Explanation of Solution

Given information:

Height of one ball $y=-16t^2+80t+25$

Height of one ball $y=-16t^2+30t+100$

Calculation:

Compute the height of two balls when they are at equal altitudes substitute =1,5in equation (1) $y=-16t^2+80t+25$ and solve for y

  $\begin{array}{l}y=-16t^2+80t+25\\ y=-16{\left(1.5\right)}^2+80\left(1.5\right)+25\\ y=109\end{array}$

Therefore, the equal height of two balls is equal to $109$ feet.