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October 22, 2014

October 22, 2014

Posted by **Anonymous** on Monday, October 24, 2011 at 9:14pm.

A projectile is fired from a cliff 100 ft above the water at an inclination of 45 degrees to the horizontal, with a muzzle velocity of 260 ft/second. The height of the projectile above the water is given by h(x)= ((-32x^2)/(260)^2)+x+100, where x is the horizontal distance of the projectile from the base of the cliff. How far from the base of the cliff is the height of the projectile a maximum?

- College Algebra -
**Reiny**, Monday, October 24, 2011 at 10:04pmThe way you typed it, the equation reduces to

h(x) = (-2/4225)x^2 + x + 100

h'(x) = (-4/4225)x + 1 = 0 for a max height

-4x = -4225

x = 1056.25 ft from base of cliff

evaluate h(1056.25) to get the actual max height.

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