Let's solve the given problem step-by-step to find the time it takes for the ball to reach the ground.
The height function for an object in free-fall can be given by the equation:
[tex]\[ h(t) = h_0 + v_0 t + \frac{1}{2} g t^2 \][/tex]
Where:
- [tex]\( h(t) \)[/tex] is the height at time [tex]\( t \)[/tex].
- [tex]\( h_0 = 480 \)[/tex] feet (initial height).
- [tex]\( v_0 = -16 \)[/tex] feet/second (initial velocity, negative because it is thrown downwards).
- [tex]\( g = -32 \)[/tex] feet/second[tex]\(^2\)[/tex] (acceleration due to gravity, negative since it is directed downwards).
However, we are given a different representation of this equation:
[tex]\[ \left(\frac{1}{2}\right) 32 t^2 + 16 t = 480 \][/tex]
Next, we recognize that this equation is equivalent to the height equation where the height reaches 0 (ground level), so we rewrite it as:
[tex]\[ 16 t^2 + 16 t - 480 = 0 \][/tex]
This is a quadratic equation in the form:
[tex]\[ at^2 + bt + c = 0 \][/tex]
Here, [tex]\( a = 16 \)[/tex], [tex]\( b = 16 \)[/tex], and [tex]\( c = -480 \)[/tex].
To solve a quadratic equation, we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ t = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 16 \cdot (-480)}}{2 \cdot 16} \][/tex]
[tex]\[ t = \frac{-16 \pm \sqrt{256 + 30720}}{32} \][/tex]
[tex]\[ t = \frac{-16 \pm \sqrt{30976}}{32} \][/tex]
[tex]\[ t = \frac{-16 \pm 176}{32} \][/tex]
This results in two possible solutions for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{160}{32} = 5 \, \text{seconds} \][/tex]
and
[tex]\[ t = \frac{-192}{32} = -6 \, \text{seconds} \][/tex]
Since time cannot be negative, we discard [tex]\( t = -6 \)[/tex] seconds.
Therefore, the correct time for the ball to reach the ground is:
[tex]\[ t = 5 \, \text{seconds} \][/tex]
So, the correct answer is:
B. 5 seconds
