Question

S=-1/2gt^2 + v_0t +s with g=32 ft/sec or g=m/sec
A bad punter on a football team kicks a football approximately straight upward with an initial velocity of 80 ft/sec
(A) If the ball leaves his foot from a height of 5 ft write an equation for the vertical height & (in ft) of the ball & seconds after being kicked.
(b) Find the time(s) at which the ball is at a height of 49 R. Round to 1 decimal place,

Answer 1

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 \( c = 11 \):

1. Calculate the discriminant \( b^2 - 4ac \):

\[
(-20)^2 - 4(4)(11) = 400 - 176 = 224
\]

2. Now, plug the values into the quadratic formula:

\[
t = \frac{-(-20) \pm \sqrt{224}}{2(4)} = \frac{20 \pm \sqrt{224}}{8}
\]

3. Calculate \( \sqrt{224} \):

\[
\sqrt{224} \approx 14.966
\]

4. 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**