A ball is thrown across a playing field from a height of h = 6 ft above the ground at an angle of 45° 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 functiony = −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.) (b) Find the horizontal distance the ball has traveled when it hits the ground. (Round your answer to one decimal place.

Question

A ball is thrown across a playing field from a height of h = 6 ft above the ground at an angle of 45° 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 functiony = −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.)  (b) Find the horizontal distance the ball has traveled when it hits the ground. (Round your answer to one decimal place.

anonymous · Answer

b) for horizontal distance, y=0

So, y = − 32/(20)^2x^2 + x + 6  =>0 = − 32/(20)^2x^2 + x + 6  ,

Solving the equation we get , x=16.9 ft , is the horizontal distance.

anonymous · Answer

how did u solve for x?

anonymous · Answer

you solved the x by using quadratic equation
0=-0.08 x^2 +x +6   x=-4.43, x=16.93 ft, disregard the x=-4.43
use x= 16.93 ft

anonymous · Answer

to find the max height use derivative of y and equate it to ZERO,to find x,...., you get x=6.25 sec
 then sub it to y=  −32/((20)^2)x^2 + x + 6 
y max= 9.125 ft6