Question

A ball is thrown across a playing field from a height of h = 5 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 function
y = −
32
(20)2
x2 + x + 5
where x is the distance in feet that the ball has traveled horizontally.

Answer 1

dy/dx = -64x/400 + 1 = 0
x = 400/64
= 25/4

Answer 2

now put 25/4 into the original equation

Answer 3

ok

Answer 4

if it doesn't work show me what you did

Answer 5

and i can walk you through

Answer 6

oops i read hight of 6ft not 5

Answer 7

oh ok i got it but now i need the horizontal distance ߺ

Answer 8

a)
differentiate
y ' = 1 - (4/25) x
max is at x such that y ' = 0
1 - (4/25) x = 0 → x = 25/4
after x = 6¼ ft the ball reaches its max height (substitute x = 25/4 in the formula which gives y)
y = 8 1/8 ft

(b)
the ball hits the ground when its height is zero
y = 0
(− 32/(20^2)) x^2 + x + 5 = 0
2x^2 - 25x - 125 = 0

x = (25 + √(25^2 - 4·2·(-125)))/4 ~ 16.3 ft

Answer 9

b is the horizontal distance

Answer 10

wait where does dx/dy come from in part 1?

Answer 11

i mean dy/dx

Answer 12

???

Answer 13

i just the way i learned it

Answer 14

its*

Answer 15

its not important

Answer 16

i mean how did you differentiate in part a?