Ask a New Question
Menu

A carnival game requires that two players stand five ft apart and throw a ball toward each other. If they are able to make the balls collide, they win. The height reached by Elios ball follows a trajectory modeled by y=-x^2 +3x+6, and the height reached by Jhumpas ball follows a trajectory modeled by y=-(x-3)^2 +9

At what height do the balls collide?
1)9ft
2)8ft
3)2ft
4)8.25ft

1 answer

To find the height at which the balls collide, we need to set the two equations equal to each other since collision occurs when they are at the same height.

The equations for the heights of the balls are:

  1. Elios's ball: \( y = -x^2 + 3x + 6 \)
  2. Jhumpas's ball: \( y = -(x - 3)^2 + 9 \)

First, we need to simplify Jhumpas's equation:

\[ y = -(x - 3)^2 + 9 \] \[ = - (x^2 - 6x + 9) + 9 \] \[ = -x^2 + 6x - 9 + 9 \] \[ = -x^2 + 6x \]

Now we have:

  1. \( y = -x^2 + 3x + 6 \)
  2. \( y = -x^2 + 6x \)

Setting these two equations equal to find the \( x \) values where they collide:

\[ -x^2 + 3x + 6 = -x^2 + 6x \]

We can eliminate \(-x^2\) from both sides:

\[ 3x + 6 = 6x \]

Rearranging gives:

\[ 6 = 6x - 3x \] \[ 6 = 3x \]

Dividing both sides by 3:

\[ x = 2 \]

Now we can substitute \( x = 2 \) back into either equation to find the height at which they collide. We'll use Elios's equation:

\[ y = -x^2 + 3x + 6 \]

Substituting \( x = 2 \):

\[ y = -(2)^2 + 3(2) + 6 \] \[ = -4 + 6 + 6 \] \[ = 8 \]

Thus, the height at which the balls collide is 8 ft.

The correct answer is:

2) 8 ft.