To determine how long it will take for the ball to hit the ground, we need to find the time [tex]\( t \)[/tex] when the height [tex]\( h(t) \)[/tex] becomes zero. The height of the ball as a function of time is given by the quadratic equation:
[tex]\[ h(t) = -16t^2 + 32t + 128 \][/tex]
The ball hits the ground when the height [tex]\( h(t) \)[/tex] is zero:
[tex]\[ -16t^2 + 32t + 128 = 0 \][/tex]
To solve this quadratic equation for [tex]\( t \)[/tex], we can use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = -16 \)[/tex], [tex]\( b = 32 \)[/tex], and [tex]\( c = 128 \)[/tex].
First, calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 32^2 - 4(-16)(128) \][/tex]
[tex]\[ \Delta = 1024 + 8192 \][/tex]
[tex]\[ \Delta = 9216 \][/tex]
Next, take the square root of the discriminant:
[tex]\[ \sqrt{\Delta} = \sqrt{9216} = 96 \][/tex]
Now, substitute the values into the quadratic formula:
[tex]\[ t = \frac{-32 \pm 96}{2(-16)} \][/tex]
This gives us two solutions:
[tex]\[ t = \frac{-32 + 96}{-32} \][/tex]
[tex]\[ t = \frac{64}{-32} \][/tex]
[tex]\[ t = -2 \][/tex]
and
[tex]\[ t = \frac{-32 - 96}{-32} \][/tex]
[tex]\[ t = \frac{-128}{-32} \][/tex]
[tex]\[ t = 4 \][/tex]
Since time cannot be negative, the valid solution is:
[tex]\[ t = 4 \][/tex]
Therefore, it will take 4 seconds for the ball to hit the ground. The correct answer is:
B. 4 seconds
