The 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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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?
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1 answer