To determine the time it takes for the ball to hit the floor, we need to find the value of \( t \) when the height \( H(t) \) equals zero. The height can be modeled by the quadratic equation:
[tex]\[ H(t) = -16t^2 + 15t + 4 \][/tex]
We set \( H(t) = 0 \):
[tex]\[ -16t^2 + 15t + 4 = 0 \][/tex]
This is a quadratic equation of the form \( at^2 + bt + c = 0 \), where:
- \( a = -16 \)
- \( b = 15 \)
- \( c = 4 \)
To solve for \( t \), we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, we compute the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
Substitute \( a = -16 \), \( b = 15 \), and \( c = 4 \) into the discriminant formula:
[tex]\[ \text{Discriminant} = 15^2 - 4(-16)(4) \][/tex]
[tex]\[ \text{Discriminant} = 225 + 256 \][/tex]
[tex]\[ \text{Sciminant} = 481 \][/tex]
Next, we substitute the values of \( a \), \( b \), and the discriminant back into the quadratic formula:
[tex]\[ t = \frac{-15 \pm \sqrt{481}}{2(-16)} \][/tex]
Calculating both potential solutions:
[tex]\[ t_1 = \frac{-15 + \sqrt{481}}{-32} \][/tex]
[tex]\[ t_2 = \frac{-15 - \sqrt{481}}{-32} \][/tex]
These yield the approximate solutions:
[tex]\[ t_1 \approx -0.217 \][/tex]
[tex]\[ t_2 \approx 1.154 \][/tex]
Given that time cannot be negative, we discard the negative solution. Therefore, the time it takes for the ball to hit the floor is approximately:
[tex]\[ t \approx 1.154 \][/tex]
So the closest answer to the options provided is:
1.15 seconds
