Answer:
The projectile will reach a height of 96 feet after about 0.84 seconds as well as after about 7.16 seconds.
Step-by-step explanation:
The height of a projectile fired upward is given by the formula:
[tex]\displaystyle s = v_{0} t - 16t^2[/tex]
Where s is the height in feet, v₀ is the initial velocity, and t is the time in seconds.
Given a projectile with an initial velocity of 128 ft/s, we want to determine how long it will take the projectile to reach a height of 96 feet.
In other words, given that v₀ = 128, find t such that s = 96.
Substitute:
[tex](96) = (128)t-16t^2[/tex]
This is a quadratic. First, we can divide both sides by -16:
[tex]-6 = -8t+t^2[/tex]
Isolate the equation:
[tex]t^2 - 8t + 6 = 0[/tex]
The equation isn't factorable, so we can consider using the quadratic formula:
[tex]\displaystyle t = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}[/tex]
In this case, a = 1, b = -8, and c = 6. Substitute:
[tex]\displaystyle t = \frac{-(-8)\pm\sqrt{(-8)^2-4(1)(6)}}{2(1)}[/tex]
Simplify:
[tex]\displaystyle t = \frac{8\pm\sqrt{40}}{2} = \frac{8\pm 2\sqrt{10}}{2} = 4\pm \sqrt{10}[/tex]
Hence, our two solutions are:
[tex]\displaystyle t = 4+\sqrt{10} \approx 7.16\text{ or } t= 4-\sqrt{10} \approx 0.84[/tex]
So, the projectile will reach a height of 96 feet after about 0.84 seconds as well as after about 7.16 seconds.