For this case we have the following quadratic equation:
\(s (t) = - 16t ^ 2 + 35t + 3\)
Where,
t: time
s: height of the ball
By the time the height is 20 feet we have:
\(-16t ^ 2 + 35t + 3 = 20\)
Rewriting the polynomial we have:
\(-16t ^ 2 + 35t + 3-20 = 0
-16t ^ 2 + 35t-17 = 0\)
Using the quadratic formula we have:
\(t = \frac{-b+/- \sqrt{b^2 - 4ac}}{2a}\)
Substituting values we have:
\(t = \frac{-35+/- \sqrt{35^2 - 4(-16)(-17)}}{2(-16)}\)
Doing the calculations we have the roots are:
\(t1 = 0.728
t2 = 1.46\)
Answer:
the time when the football will be 20ft above the ground is:
B) 0.73 seconds or 1.46 seconds
