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A projectile is launched upward. The function f(x)=-16x2 + 64x describe the position of the projectiles height, in feet, x seconds after it is launched.

A Projectile Is Launched Upward The Function Fx16x2 64x Describe The Position Of The Projectiles Height In Feet X Seconds After It Is Launched class=

Sagot :

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.

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