To determine how long it takes for the tennis ball to hit the ground, we need to find when the height of the ball is 0.
The quadratic function modeling the height of the ball is given by:
[tex]\[ p(t) = -16t^2 + 46t + 6 \][/tex]
To find when the ball hits the ground, we need to solve for [tex]\( t \)[/tex] when [tex]\( p(t) = 0 \)[/tex]:
[tex]\[ -16t^2 + 46t + 6 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex]. For our equation, the coefficients are as follows:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 46 \)[/tex]
- [tex]\( c = 6 \)[/tex]
Solving a quadratic equation typically involves using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Let's identify the solutions of this equation:
[tex]\[ t = \frac{-46 \pm \sqrt{46^2 - 4(-16)(6)}}{2(-16)} \][/tex]
This will provide us with two potential values for [tex]\( t \)[/tex], representing the times at which the ball reaches the ground. For a clear understanding, the discriminant [tex]\( b^2 - 4ac \)[/tex] inside the square root needs to be calculated and must be positive for real solutions. Upon solving, we would find two values, [tex]\( t = -\frac{1}{8} \)[/tex] and [tex]\( t = 3 \)[/tex].
Since time cannot be negative, we disregard [tex]\( t = -\frac{1}{8} \)[/tex].
Thus, the time it takes for the ball to hit the ground is:
[tex]\[ t = 3 \text{ seconds} \][/tex]
In conclusion, it takes 3 seconds for the ball to hit the ground.
