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(x)=(-32x^2)/((170)^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?
:
Convert the coefficient of x^2 to a more manageable value
:
{{{-32/10^2}}} = -.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 = {{{(-1)/(2*-.001107)}}}
x = +451.6 ft from the base of the cliff, it will be at max height