A ball is thrown across a playing field from a height of h

A ball is thrown across a playing field from a height of h = 6 ft above the ground at an angle of 45 degree to the horizontal at the speed of 20 ft/s. It can be deduced from physical principles that the path of the ball is modeled by the function y = - 32/(20)^2 x^2 + x + 6 where x is the distance in feet that the ball has traveled horizontally. Find the maximum height attained by the ball. (Round your answer to three decimal places.) Find the horizontal distance the ball has traveled when it hits the ground. ()

Solution

a)y =-(32/20²)x² +x+6

y =-(32/400)x² +x+6

for maximum height dy/dx =0

dy/dx=-(32/400)2x +1+0=0

-(32/200)x =-1

x=200/32

x=6.25

maximum height =y =-(32/400)6.25² +6.25+6

maximum height =y=9.125 ft

b) hits the ground =>y=0

-(32/400)x² +x+6=0

-(4/50)x² +x+6=0

-4x² +50x+300=0

x=[-50+(50²-4*(-4)*300)]/2(-4) ,x=[-50-(50²-4*(-4)*300)]/2(-4)

x=-4.43 ,x=16.93

x cannot be negative

horizontal distance ball travelled =16.93ft before it hits the ground