Question

A projectile is fired from a cliff 210 feet above the water at an inclination of 45 degree to the horizontal, with a muzzle velocity of 45 feet per second. The height h of the projectile above the water is given by h(x) = -32x^2/(45)^2 + x +210 where x is the horizontal distance of the projectile from the face of the cliff. Use this information to answer the following.

At what horizontal distance from the face of the cliff is the height of the projectile a maximum? x = _____ feet

Answer 1

31.64 feet

Explanation:

The height is given by the equation:

`h(x) = -32x^2/45^2 + x +210 = -(32)/(45^2)x^2 + x +210`

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
`x^n` is
`nx^(n-1)`, we have:

`h'(x) = (-(32)/(45^2)x^2 + x +210)' = (-(32)/(45^2)x^2)' + (x)' +(210)'=-2(32)/(45^2)x+1`

And we want h'(x)=0, so:

`-2(32)/(45^2)x+1=0`

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:

`2(32)/(45^2)x=1`

`x=(1)/(2)(45^2)/(32)=31.64`