Question

1. a frisbee is thrown into the air with an initial velocity of 40 feet per second. the release point is 5 feet above the ground. the function ( h=-16t^{2}+40t + 5 ) represents the height ( h ) in feet of the frisbee after ( t ) seconds.

a) find the height of the frisbee each second it is in the air.
b) use the model to determine how long the frisbee is in the air.
□ optional: complete problems 1 - 6 from your textbook on separate lined paper.
when you are finished, check the solutions at the back of the textbook.

Explanation:

Step1: Calculate height at each second for part a)

• For \(t = 0\):

\(h=-16(0)^{2}+40(0)+5 = 5\)

• For \(t = 1\):

\(h=-16(1)^{2}+40(1)+5=-16 + 40+5=29\)

• For \(t = 2\):

\(h=-16(2)^{2}+40(2)+5=-64 + 80+5 = 21\)

• For \(t = 3\):

\(h=-16(3)^{2}+40(3)+5=-144+120 + 5=-19\) (Since height can't be negative, we ignore \(t = 3\) as the frisbee has already landed before \(t=3\))

Step2: Solve for when \(h = 0\) for part b)

Set \(h=-16t^{2}+40t + 5=0\)
Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) where \(a=-16\), \(b = 40\), \(c = 5\)
\(t=\frac{-40\pm\sqrt{40^{2}-4(-16)(5)}}{2(-16)}=\frac{-40\pm\sqrt{1600 + 320}}{-32}=\frac{-40\pm\sqrt{1920}}{-32}=\frac{-40\pm8\sqrt{30}}{-32}\)
We take the positive root \(t=\frac{-40 + 8\sqrt{30}}{-32}\approx\frac{-40+43.82}{-32}\) (rejected) or \(t=\frac{-40-8\sqrt{30}}{-32}\approx\frac{-40 - 43.82}{-32}\approx2.62\)

Answer:

a) At \(t = 0\), height \(h = 5\) feet; at \(t = 1\), height \(h = 29\) feet; at \(t = 2\), height \(h=21\) feet.
b) The frisbee is in the air for approximately \(t\approx2.62\) seconds.