Question

1. the canyon archer

an archer standing on the rim of a canyon shoots an arrow upward with an initial velocity of 65 feet per second.
the rim of the canyon is 250 feet above the canyon floor. the height of the arrow is modeled by
( h(t) = -16t^2 + 65t + 250 ). to the nearest hundredth of a second, how many seconds pass before the arrow is 100 feet below its starting position?
sketch:
algebraically
( 100 ft = -16t^2 + 65t + 250 )
( -100 ) ( -100 )
( -16t^2 + 65t + 150 )

Explanation:

Step1: Define target height

The starting position is 250 ft, so 100 ft below is $250 - 100 = 150$ ft. Set $h(t)=150$.
$$150 = -16t^2 + 65t + 250$$

Step2: Rearrange to standard quadratic form

Subtract 150 from both sides to set equation to 0.
$$-16t^2 + 65t + 100 = 0$$
Multiply by -1 for simplicity:
$$16t^2 - 65t - 100 = 0$$

Step3: Apply quadratic formula

For $at^2+bt+c=0$, $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=16$, $b=-65$, $c=-100$.
$$t=\frac{65\pm\sqrt{(-65)^2-4(16)(-100)}}{2(16)}$$

Step4: Calculate discriminant

Compute the value under the square root.
$$\sqrt{4225 + 6400} = \sqrt{10625} \approx 103.0776$$

Step5: Solve for positive t

Only positive time is valid, so use the plus sign.
$$t=\frac{65 + 103.0776}{32} \approx \frac{168.0776}{32} \approx 5.25$$

Answer:

5.25 seconds