A cannonball fred out to sea from a shore battery follows a

A cannonball fred out to sea from a shore battery follows a parabolic trajectory given by the graph of the equat h(x) = 8x-0.0 1x2 where h(x) is the height of the cannonball above the water when it has traveled a horizontal distance of x feet. hCr) (a) What is the maximum height that the cannonball reaches? 1600 ft b) How far does the cannonball travel horizontally before splashing into the water? ft

Solution

h(x)=8x-0.01x²

a. x=-b/2a =-(8)/(2*-0.01) = 400 ft

h(400)= 8(400)-0.01(400)²=1600 ft .

b. 0= 8x-0.01x²

x(8-0.01x)=0

x=0, x=800

So it travels 800 ft before hitting the water .