To determine how long it takes for the arrow to reach a height of 48 feet given an initial velocity of 96 feet per second, we start with the equation for height:
[tex]\[ s = v_0 t - 16 t^2 \][/tex]
Given:
- Initial velocity, [tex]\( v_0 = 96 \)[/tex] ft/s
- Height, [tex]\( s = 48 \)[/tex] ft
The problem can be set up as:
[tex]\[ 48 = 96 t - 16 t^2 \][/tex]
Rearranging it into a standard quadratic form:
[tex]\[ 16 t^2 - 96 t + 48 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex] where:
- [tex]\( a = 16 \)[/tex]
- [tex]\( b = -96 \)[/tex]
- [tex]\( c = 48 \)[/tex]
We can solve this using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, we need to calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
Substituting in the values:
[tex]\[ \text{Discriminant} = (-96)^2 - 4(16)(48) \][/tex]
[tex]\[ \text{Discriminant} = 9216 - 3072 \][/tex]
[tex]\[ \text{Discriminant} = 6144 \][/tex]
The discriminant is 6144, which is a positive number, indicating there are two real roots. Now, applying the quadratic formula:
[tex]\[ t = \frac{96 \pm \sqrt{6144}}{32} \][/tex]
Solving for the roots:
[tex]\[ t_1 = \frac{96 + \sqrt{6144}}{32} \][/tex]
[tex]\[ t_1 \approx \frac{96 + 78.40}{32} \][/tex]
[tex]\[ t_1 \approx \frac{174.40}{32} \][/tex]
[tex]\[ t_1 \approx 5.45 \][/tex]
[tex]\[ t_2 = \frac{96 - \sqrt{6144}}{32} \][/tex]
[tex]\[ t_2 \approx \frac{96 - 78.40}{32} \][/tex]
[tex]\[ t_2 \approx \frac{17.60}{32} \][/tex]
[tex]\[ t_2 \approx 0.55 \][/tex]
Thus, the arrow reaches a height of 48 feet at approximately [tex]\( t_1 = 5.45 \)[/tex] seconds and [tex]\( t_2 = 0.55 \)[/tex] seconds, when rounded to the nearest hundredth.
