Question

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. Isaac throws his racquet toward it at a trajectory modeled by y=
3
1
​
x+4. When does his racquet hit the ball?
A. after 8.4 milliseconds
B. after 1.5 milliseconds
C. after 6.8 milliseconds
D. after 3.5 milliseconds

Answer 1

To find when Isaac's racquet hits the ball, we need to set the two equations equal to each other since they represent the same height at the same time.

1. The height of the ball is given by:
\[
y = -3|x - 4| + 20
\]

2. The height of Isaac's racquet is given by:
\[
y = \frac{3}{1}x + 4 = 3x + 4
\]

Setting the two equations equal:
\[
-3|x - 4| + 20 = 3x + 4
\]

We will solve this equation by considering the two cases of the absolute value.

### Case 1: \(x - 4 \geq 0\) (i.e., \(x \geq 4\))
In this case, \(|x - 4| = x - 4\). Thus, the equation becomes:
\[
-3(x - 4) + 20 = 3x + 4
\]
Simplifying:
\[
-3x + 12 + 20 = 3x + 4
\]
\[
32 - 3x = 3x + 4
\]
\[
32 - 4 = 6x
\]
\[
28 = 6x
\]
\[
x = \frac{28}{6} = \frac{14}{3} \approx 4.67
\]

### Case 2: \(x - 4 < 0\) (i.e., \(x < 4\))
In this case, \(|x - 4| = -(x - 4) = 4 - x\). Now we solve:
\[
-3(4 - x) + 20 = 3x + 4
\]
Simplifying:
\[
-12 + 3x + 20 = 3x + 4
\]
\[
8 = 3x + 4
\]
\[
8 - 4 = 3x
\]
\[
4 = 3x
\]
\[
x = \frac{4}{3} \approx 1.33
\]

### Summary of Solutions
From the two cases, we have:
- From Case 1: \(x \approx 4.67\)
- From Case 2: \(x \approx 1.33\)

### Conclusion
Isaac's racquet hits the ball at approximately \(1.33\) milliseconds, which matches option B:

**B. after 1.5 milliseconds.**

Note that the closest option given for \(1.33\) in the choices is \(1.5\).