analyzing the motion of a projectile: a projectile is fired from a cliff 200 feet above the water at an inclination of 45 degrees, with a muzzle velocity of 50 feet per second. the height of the projectile above the water is given by h(x)= -0.0128x^2+x+200 where x is the horizontal distance of the projectile from the base of the cliff.

Question

a) how far from the base of the cliff is the height of the projectile at its maximum?
b) find the maximum height of the projectile.
c) how far from the base of the cliff will the projectile strike the water?
d) when the height of the projectile is 100 feet above the water, how far is it horizontally from the cliff?

Answer 1

This is just a good old parabola.

max height occurs at x=-b/2a = -1/-.0256 = 39.0625

h(39.0625) = 219.531

h(x) = 0 at x = 170 approx.

h(x) = 100 at x = 135.7

Answer 2

thanks! i posted some other questions that i really need help with too