To determine how high the cannonball travels before it begins to fall back to the ground, we need to find the maximum height reached by the cannonball. Let's break down the steps to solve this problem:
1. Identify the given information:
- The initial velocity [tex]\( v_0 \)[/tex] is 420 feet per second.
- The equation for height [tex]\( h \)[/tex] as a function of time [tex]\( t \)[/tex] is given by:
[tex]\[
h(t) = -16t^2 + 420t
\][/tex]
2. Determine the time at which the cannonball reaches its maximum height:
The maximum height occurs at the peak of the projectile's path. To find this, we need the vertex of the parabolic equation. For a quadratic equation in the form [tex]\( h(t) = at^2 + bt + c \)[/tex], the time [tex]\( t \)[/tex] at which the maximum height occurs is given by:
[tex]\[
t = -\frac{b}{2a}
\][/tex]
Here, [tex]\( a = -16 \)[/tex] and [tex]\( b = 420 \)[/tex]. Plugging in these values, we get:
[tex]\[
t = -\frac{420}{2(-16)} = \frac{420}{32} = 13.125 \text{ seconds}
\][/tex]
3. Calculate the maximum height:
Substitute the time [tex]\( t = 13.125 \)[/tex] back into the height equation to determine the maximum height [tex]\( h \)[/tex]:
[tex]\[
h(13.125) = -16 \times (13.125)^2 + 420 \times 13.125
\][/tex]
Let's evaluate this step by step:
[tex]\[
13.125^2 = 172.265625
\][/tex]
[tex]\[
-16 \times 172.265625 = -2756.25
\][/tex]
[tex]\[
420 \times 13.125 = 5512.5
\][/tex]
Now, combine these values:
[tex]\[
h(13.125) = -2756.25 + 5512.5 = 2756.25 \text{ feet}
\][/tex]
So, the maximum height the cannonball reaches before it begins to fall back to the ground is [tex]\( 2756.25 \)[/tex] feet.
