Answer:
31.64 feet
Step-by-step explanation:
The height is given by the equation:
[tex]h(x) = -32x^2/45^2 + x +210 = -\frac{32}{45^2}x^2 + x +210[/tex]
where x is the horizontal distance of the projectile from the face of the cliff.
If the height is maximum, then its derivative must be h'(x)=0. Remembering that the derivative of [tex]x^n[/tex] is [tex]nx^{n-1}[/tex], we have:
[tex]h'(x) = (-\frac{32}{45^2}x^2 + x +210)' = (-\frac{32}{45^2}x^2)' + (x)' +(210)'=-2\frac{32}{45^2}x+1[/tex]
And we want h'(x)=0, so:
[tex]-2\frac{32}{45^2}x+1=0[/tex]
This will give us the horizontal distance from the face of the cliff when the height of the projectile is at its maximum. We then do:
[tex]2\frac{32}{45^2}x=1[/tex]
[tex]x=\frac{1}{2}\frac{45^2}{32}=31.64[/tex]
