To determine if the firecracker will reach a height of 100 feet before it bursts, we'll use the given height equation:
[tex]\[ h = -16t^2 + vt + c \][/tex]
where:
- [tex]\( v \)[/tex] is the initial velocity,
- [tex]\( t \)[/tex] is the time in seconds,
- [tex]\( c \)[/tex] is the initial height above the ground.
Given the parameters:
- Initial velocity, [tex]\( v = 100 \)[/tex] feet per second,
- Initial height (height of the building), [tex]\( c = 15 \)[/tex] feet,
- Target height, [tex]\( h = 100 \)[/tex] feet.
The height equation becomes:
[tex]\[ 100 = -16t^2 + 100t + 15 \][/tex]
To find the time [tex]\( t \)[/tex] when the height [tex]\( h \)[/tex] is 100 feet, we rearrange this equation to:
[tex]\[ -16t^2 + 100t + 15 - 100 = 0 \][/tex]
[tex]\[ -16t^2 + 100t - 85 = 0 \][/tex]
This is a quadratic equation in the standard form [tex]\( at^2 + bt + c = 0 \)[/tex], where:
- [tex]\( a = -16 \)[/tex],
- [tex]\( b = 100 \)[/tex],
- [tex]\( c = -85 \)[/tex].
To solve this quadratic equation, we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 100^2 - 4(-16)(-85) \][/tex]
[tex]\[ \Delta = 10000 - 5440 \][/tex]
[tex]\[ \Delta = 4560 \][/tex]
Since the discriminant ([tex]\(\Delta\)[/tex]) is greater than zero, there are two real solutions for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-b \pm \sqrt{4560}}{2(-16)} \][/tex]
Now calculate the specific solutions:
[tex]\[ t_1 = \frac{-100 + \sqrt{4560}}{-32} \][/tex]
[tex]\[ t_1 \approx \frac{-100 + 67.54}{-32} \][/tex]
[tex]\[ t_1 \approx \frac{-32.46}{-32} \][/tex]
[tex]\[ t_1 \approx 1.015 \][/tex]
[tex]\[ t_2 = \frac{-100 - \sqrt{4560}}{-32} \][/tex]
[tex]\[ t_2 \approx \frac{-100 - 67.54}{-32} \][/tex]
[tex]\[ t_2 \approx \frac{-167.54}{-32} \][/tex]
[tex]\[ t_2 \approx 5.235 \][/tex]
Both [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex] are positive, meaning the firecracker will reach the height of 100 feet at two different times:
1. When [tex]\( t \approx 1.015 \)[/tex] seconds (initial ascent),
2. When [tex]\( t \approx 5.235 \)[/tex] seconds (descent).
Using the max of these times, [tex]\( t \approx 5.235 \)[/tex] seconds.
Therefore, the firecracker will indeed reach a height of 100 feet before it bursts. This means the quadratic equation solutions confirm that it will reach 100 feet as it moves upward. So, the correct explanation is that the firecracker will reach 100 feet, specifically at around 1.015 seconds and 5.235 seconds after being launched.
