Question

1. in a circus, a clown throws an apple up off a platform. the height of the apple vs. time can be given by the following equation: $y = -x^{2} + 4x + 5$. another clown tries to shoot the apple with a bow and arrow. the height of the arrow vs. time can be given by the following equation: $y = \frac{7}{3}x + 1$. when will they make contact with each other?

Explanation:

Step1: Set equations equal

Set the height equations equal to find contact time:
$$-x^2 + 4x + 5 = \frac{7}{3}x + 1$$

Step2: Eliminate fraction, simplify

Multiply all terms by 3 to eliminate the fraction, then rearrange into standard quadratic form:
$$-3x^2 + 12x + 15 = 7x + 3$$
$$-3x^2 + 5x + 12 = 0$$
Multiply by -1 to make the leading coefficient positive:
$$3x^2 - 5x - 12 = 0$$

Step3: Solve quadratic equation

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=3$, $b=-5$, $c=-12$:
$$x = \frac{5 \pm \sqrt{(-5)^2 - 4(3)(-12)}}{2(3)}$$
$$x = \frac{5 \pm \sqrt{25 + 144}}{6}$$
$$x = \frac{5 \pm \sqrt{169}}{6}$$
$$x = \frac{5 \pm 13}{6}$$

Step4: Evaluate valid solution

Calculate the two solutions:
$x = \frac{5 + 13}{6} = 3$, $x = \frac{5 - 13}{6} = -\frac{4}{3}$
Discard the negative time (invalid in this context).

Answer:

They make contact at $x=3$ units of time.