Let's solve the given problem step-by-step:
First, we need to use the formula for the height of the ball:
[tex]\[ h = -16t^2 + v_0 t \][/tex]
where:
- [tex]\( h \)[/tex] is the height of the ball,
- [tex]\( t \)[/tex] is the time in seconds,
- [tex]\( v_0 \)[/tex] is the initial velocity.
We are given:
- The initial velocity [tex]\( v_0 = 96 \)[/tex] feet per second,
- We need to find the time [tex]\( t \)[/tex] when the height [tex]\( h \)[/tex] is 48 feet.
Substitute the values into the formula:
[tex]\[ 48 = -16t^2 + 96t \][/tex]
Rearrange the equation to standard quadratic form:
[tex]\[ -16t^2 + 96t - 48 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex]:
[tex]\[ -16t^2 + 96t - 48 = 0 \][/tex]
To solve for [tex]\( t \)[/tex], we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In our equation:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 96 \)[/tex]
- [tex]\( c = -48 \)[/tex]
Plug these values into the quadratic formula:
[tex]\[ t = \frac{-96 \pm \sqrt{96^2 - 4(-16)(-48)}}{2(-16)} \][/tex]
Simplify inside the square root:
[tex]\[ t = \frac{-96 \pm \sqrt{9216 - 3072}}{-32} \][/tex]
[tex]\[ t = \frac{-96 \pm \sqrt{6144}}{-32} \][/tex]
Calculate the square root:
[tex]\[ t = \frac{-96 \pm 78.4}{-32} \][/tex]
We have two potential solutions for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-96 + 78.4}{-32} \][/tex]
[tex]\[ t = \frac{-96 - 78.4}{-32} \][/tex]
Solve these individually:
For [tex]\( t = \frac{-96 + 78.4}{-32} \)[/tex]:
[tex]\[ t = \frac{-17.6}{-32} \][/tex]
[tex]\[ t \approx 0.55 \][/tex]
For [tex]\( t = \frac{-96 - 78.4}{-32} \)[/tex]:
[tex]\[ t = \frac{-174.4}{-32} \][/tex]
[tex]\[ t \approx 5.45 \][/tex]
These are the values when the ball's height will be 48 feet. Rounding to the nearest tenth of a second, we get:
[tex]\[ t \approx 0.6 \text{ seconds and } 5.4 \text{ seconds} \][/tex]
So, the ball reaches a height of 48 feet at approximately [tex]\( 0.6 \)[/tex] seconds and [tex]\( 5.4 \)[/tex] seconds after it is thrown.
