To solve the problem involving the projectile fired straight up with a velocity of [tex]\( 368 \)[/tex] ft/s, we will use the given height equation:
[tex]\[ h(t) = -16t^2 + 368t. \][/tex]
### Part (a): Finding the Maximum Height
The height equation [tex]\( h(t) = -16t^2 + 368t \)[/tex] is a quadratic equation of the form [tex]\( h(t) = at^2 + bt + c \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 368 \)[/tex], and [tex]\( c = 0 \)[/tex].
For a quadratic equation [tex]\( at^2 + bt + c \)[/tex], the vertex form, which gives the maximum (or minimum) point, occurs at [tex]\( t = -\frac{b}{2a} \)[/tex].
Plugging in the values [tex]\( a = -16 \)[/tex] and [tex]\( b = 368 \)[/tex]:
[tex]\[ t = -\frac{368}{2 \times -16} = -\frac{368}{-32} = 11.5 \text{ seconds}. \][/tex]
This is the time at which the projectile reaches its maximum height.
To find the maximum height, substitute [tex]\( t = 11.5 \)[/tex] back into the height equation:
[tex]\[ h(11.5) = -16(11.5)^2 + 368(11.5). \][/tex]
[tex]\[ h(11.5) = -16 \times 132.25 + 368 \times 11.5. \][/tex]
[tex]\[ h(11.5) = -2116 + 4232. \][/tex]
[tex]\[ h(11.5) = 2116 \text{ ft}. \][/tex]
Therefore, the maximum height of the projectile is
[tex]\[ \boxed{2116} \text{ ft}. \][/tex]
### Part (b): Finding the Time When the Projectile Hits the Ground
The projectile hits the ground when its height equals zero. In other words, we need to solve for [tex]\( t \)[/tex] when [tex]\( h(t) = 0 \)[/tex].
So, set the height equation to zero:
[tex]\[ 0 = -16t^2 + 368t. \][/tex]
This is a quadratic equation in standard form. We can factor out [tex]\( t \)[/tex]:
[tex]\[ 0 = t(-16t + 368). \][/tex]
Setting each factor equal to zero gives us:
[tex]\[ t = 0 \quad \text{or} \quad -16t + 368 = 0. \][/tex]
Solve for [tex]\( t \)[/tex] in [tex]\(-16t + 368 = 0\)[/tex]:
[tex]\[ -16t + 368 = 0. \][/tex]
[tex]\[ -16t = -368. \][/tex]
[tex]\[ t = \frac{368}{16} = 23 \text{ seconds}. \][/tex]
Thus, the projectile hits the ground at [tex]\( t = 23 \)[/tex] seconds.
Therefore, the time when the projectile hits the ground is
[tex]\[ \boxed{23} \text{ seconds}. \][/tex]
### Summary
The maximum height the projectile reaches is [tex]\( \boxed{2116} \)[/tex] ft. The time at which the projectile hits the ground is [tex]\( \boxed{23} \)[/tex] seconds.
