Question

1. modeling real life you throw a softball straight up into the air with an initial vertical velocity of 40 feet per second. the release point is 5 feet above the ground. the function ( h = - 16 t ^ { 2 } + 40 t + 5 ) represents the height ( h ) (in feet) of the softball after ( t ) seconds.

a. find the height of the softball each second after it is released.
b. estimate when the height of the softball is 15 feet.
c. using a graph, after how many seconds is the softball 15 feet above the ground?

Explanation:

Step1: Calculate height for each second (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 in real - life context for this problem after throwing, we consider non - negative values. The ball hits the ground before \(t = 3\))

Step2: Solve for \(t\) when \(h = 15\) (part b and c)

Set \(h = 15\), so \(-16t^2+40t + 5=15\)
\(-16t^2+40t-10 = 0\)
Multiply through by \(- 2\) to get \(8t^2-20t + 5=0\)
Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) where \(a = 8\), \(b=-20\), \(c = 5\)
\(t=\frac{20\pm\sqrt{(-20)^{2}-4\times8\times5}}{2\times8}=\frac{20\pm\sqrt{400 - 160}}{16}=\frac{20\pm\sqrt{240}}{16}=\frac{20\pm4\sqrt{15}}{16}=\frac{5\pm\sqrt{15}}{4}\)
\(t_1=\frac{5+\sqrt{15}}{4}\approx\frac{5 + 3.87}{4}=\frac{8.87}{4}\approx2.22\)
\(t_2=\frac{5-\sqrt{15}}{4}\approx\frac{5 - 3.87}{4}=\frac{1.13}{4}\approx0.28\)

Answer:

a. At \(t = 0\) s, \(h = 5\) ft; at \(t = 1\) s, \(h = 29\) ft; at \(t = 2\) s, \(h = 21\) ft.
b. Approximately \(t\approx0.28\) s and \(t\approx2.22\) s.
c. Approximately \(t\approx0.28\) s and \(t\approx2.22\) s.