To model the height of the football over time after being kicked, we can use the quadratic formula provided:
\[ S(t) = -\frac{1}{2}gt^2 + v_0t + s \]
where:
- \( g = 32 \text{ ft/sec}^2 \) (acceleration due to gravity),
- \( v_0 = 80 \text{ ft/sec} \) (initial velocity),
- \( s = 5 \text{ ft} \) (initial height).
(A) Vertical Height Equation
Given these values, we can substitute them into the formula:
\[ S(t) = -\frac{1}{2}(32)t^2 + 80t + 5 \]
Simplifying this, we get:
\[ S(t) = -16t^2 + 80t + 5 \]
(B) Finding the Time(s) at Which the Ball is at a Height of 49 ft
To find the time(s) when the ball is at a height of 49 ft, we set \( S(t) = 49 \):
\[ -16t^2 + 80t + 5 = 49 \]
Now, we can rearrange the equation:
\[ -16t^2 + 80t + 5 - 49 = 0 \]
\[ -16t^2 + 80t - 44 = 0 \]
To simplify, we can divide the entire equation by -4:
\[ 4t^2 - 20t + 11 = 0 \]
Now we can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 4 \), \( b = -20 \), and \( c = 11 \):
- Calculate the discriminant \( b^2 - 4ac \):
\[ (-20)^2 - 4(4)(11) = 400 - 176 = 224 \]
- Now, plug the values into the quadratic formula:
\[ t = \frac{-(-20) \pm \sqrt{224}}{2(4)} = \frac{20 \pm \sqrt{224}}{8} \]
- Calculate \( \sqrt{224} \):
\[ \sqrt{224} \approx 14.966 \]
- Continue with the formula:
\[ t = \frac{20 \pm 14.966}{8} \]
This gives us two potential times:
\[ t_1 = \frac{20 + 14.966}{8} \approx \frac{34.966}{8} \approx 4.371 \] \[ t_2 = \frac{20 - 14.966}{8} \approx \frac{5.034}{8} \approx 0.629 \]
Final Answers
So the two times at which the ball is at a height of 49 ft, rounded to one decimal place, are approximately:
- t = 4.4 seconds
- t = 0.6 seconds
2 answers