SOLUTION: A projectile is fired from a cliff 100 feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of 170 feet per second. The height h of the projecti

Algebra ->  Equations -> SOLUTION: A projectile is fired from a cliff 100 feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of 170 feet per second. The height h of the projecti      Log On


   

Question 471804: A projectile is fired from a cliff 100 feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of 170 feet per second. The height h of the projectile above the water is given by
++h%28x%29=%28-32x%5E2%29%2F%28%28170%29%5E2%29%2Bx%2B100+
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?
A. 225.78 ft
B. 451.56 ft
C. 325.78 ft
D. 777.34 ft


Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A projectile is fired from a cliff 100 feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of 170 feet per second. The height h of the projectile above the water is given by
++h%28x%29=%28-32x%5E2%29%2F%28%28170%29%5E2%29%2Bx%2B100+
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?
:
Convert the coefficient of x^2 to a more manageable value
:
-32%2F10%5E2 = -.001107, so we have
h(x) = -.001107x^2 + x + 100
:
find the axis of symmetry, x = -b/(2a), in this equation a=-.001107, b=1
x = %28-1%29%2F%282%2A-.001107%29
x = +451.6 ft from the base of the cliff, it will be at max height