Samantha and Isaac are playing racquetball. Samantha hits the ball, sending it on a trajectory modeled by y=-3|x-4|+20, where y is the height reached by the ball, in feet, after x milliseconds. In a desperate attempt to keep the ball in the air, Isaac throws his racquet toward it at a trajectory modeled by y=1/3x+4. When does his racquet hit the ball?

Question

1)after 1.5 milliseconds
2)after 6.8 milliseconds
3)after 8.4 milliseconds
4)after 3.5 milliseconds

Answer 1

To determine when Isaac's racquet hits the ball, we need to find the value of \(x\) when the two trajectories are equal. This means we have to set the two equations equal to each other:

\[ -3 |x-4| + 20 = \frac{1}{3}x + 4 \]

First, let's solve for \(x\). We will handle the absolute value by splitting it into two cases.

Case 1: \(x - 4 \geq 0\) (i.e., \(x \geq 4\))

In this case, \(|x - 4| = x - 4\), so the equation becomes: \[ -3(x - 4) + 20 = \frac{1}{3}x + 4 \] Simplifying this, \[ -3x + 12 + 20 = \frac{1}{3}x + 4 \ -3x + 32 = \frac{1}{3}x + 4 \ -3x - \frac{1}{3}x = 4 - 32 \ -\frac{9}{3}x - \frac{1}{3}x = -28 \ -\frac{10}{3}x = -28 \] Multiplying both sides by \(-\frac{3}{10}\), \[ x = \frac{-28 \cdot -3}{10} = \frac{84}{10} = 8.4 \]

Case 2: \(x - 4 < 0\) (i.e., \(x < 4\))

In this case, \(|x - 4| = -(x - 4) = 4 - x\), so the equation becomes: \[ -3(4 - x) + 20 = \frac{1}{3}x + 4 \] This simplifies to: \[ -12 + 3x + 20 = \frac{1}{3}x + 4 \ 3x + 8 = \frac{1}{3}x + 4 \] Multiplying the entire equation by 3 to eliminate the fraction: \[ 9x + 24 = x + 12 \ 9x - x = 12 - 24 \ 8x = -12 \ x = -\frac{12}{8} = -1.5 \] Since \(-1.5\) is not a valid time (it cannot be negative), we disregard this solution.

Conclusion

The only valid solution for when Isaac's racquet hits the ball is from Case 1: \[ x = 8.4 \text{ milliseconds} \]

Thus, the correct answer is: 3) after 8.4 milliseconds.