Question

try: the chs baseball team was on the field and the batter popped the ball up. the equation ( h(t) = 80t - 16t^2 + 3.5 ) represents the height of the ball above the ground in feet as a function of time in seconds. how long will the catcher have to get in position to catch the ball before it hits the ground? round to the nearest second. (hint: make sure the equation is in a, b, c order before doing anything!)

Explanation:

Step1: Set height to 0

When the ball hits the ground, \( h(t) = 0 \). So we set up the equation:
\( 0 = -16t^2 + 80t + 3.5 \) (rewriting \( 80t - 16t^2 + 3.5 \) in standard quadratic form \( at^2 + bt + c = 0 \) as \( -16t^2 + 80t + 3.5 = 0 \))

Step2: Use quadratic formula

For a quadratic equation \( ax^2 + bx + c = 0 \), the solutions are given by \( t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \). Here, \( a=-16 \), \( b = 80 \), \( c = 3.5 \).

First, calculate the discriminant \( D=b^2 - 4ac \):
\( D=(80)^2-4\times(-16)\times(3.5) \)
\( D = 6400+224 \)
\( D = 6624 \)

Then, find the roots:
\( t=\frac{-80\pm\sqrt{6624}}{2\times(-16)} \)
\( \sqrt{6624}\approx81.39 \)
So we have two solutions:
\( t_1=\frac{-80 + 81.39}{-32}=\frac{1.39}{-32}\approx - 0.043 \) (discard since time can't be negative)
\( t_2=\frac{-80 - 81.39}{-32}=\frac{-161.39}{-32}\approx5.043 \)

Step3: Round to nearest second

Rounding \( 5.043 \) to the nearest second gives \( 5 \).

Answer:

5