Given:
[tex]f(x)=-16x^2+64x[/tex]Differentiate with respect to x, we get
[tex]f^{\prime}(x)=-16\times2x+64(1)[/tex][tex]f^{\prime}(x)=-32x+64[/tex][tex]\text{Set f'(x)=0 to find the maxi}mum\text{ value of x}[/tex][tex]0=-32x+64[/tex]Adding 32x to both sides of the equation, we get
[tex]0+32x=-32x+64+32x[/tex][tex]32x=64[/tex]Dividing both sides by 32, we get
[tex]\frac{32x}{32}=\frac{64}{32}[/tex][tex]x=2[/tex]We get the maximum value of x is 2.
Substitute x=2 in the given function to find the maximum height of the projectiles.
[tex]f(2)=-16(2)^2+64\times2[/tex][tex]f(2)=64[/tex]Hence the maximum height of the projectiles is 64 feet.
Substitute f(x) =0 in the given function to find the number of seconds the rocket takes to hit the ground.
[tex]0=-16x^2+64x[/tex][tex]0=-16x(x-4)[/tex][tex]x=0\text{ or (x-4)=0}[/tex][tex]\text{Consider x-4=0.}[/tex][tex]x=4\text{ }[/tex]After 4.0 seconds, the rocket hit the ground.