Agolf ball travels a distance of 600 feet as measured along the ground and reaches an altitude of 200 feet. if the origin represents the tee and the ball travels along a parabolic path that opens downward, find an equation for the path of the golf ball.

Y = y0 + V0y * t + gt^2 / 2

y0 = 0

g  ≈ - 32 ft / s^2

(1) y =  V0y * t - 16t^2

(2) x = V0x * t 

Use (1) and the maximum heigth formula to determine V0y

y max = (V0y)^2 / (2g) = 200 => (V0y)^2 = 2*32*200 = 12,800 => V0y ≈ 113.14 ft/s

y = 113.14 t - 16t^2    

for y = 200 => 200 = 113t - 16t^2 => -16t^2 + 113.14t - 200 = 0

Solve that equation using the quadratic formula and you will ge t = 3.54 s

The total time is 3.54 s * 2 = 7.08 s

Then use that time in (2) to find V0x

600 = V0x * t => V0x = 600 / 7.08 s = V0x = 84.7 ft/s

Then, x = 84.7 t.
Now solve for t and replace it in  y = 113.14 t - 16t^2 :

t = x / 84.7

y = 113.14 (x/84.7) - 16(x^2 / 84.7 ^2) =  1.336 x - 0.00223x^2

You can check that for, except for a small difference due to the approximations, for x = 600, y = 0 and for x =300 y = 200

y = 1.336 x - 0.00223x^2

y=1.32x-0.0022

Step-by-step explanation:

Let the standard equation of parabola be y=a+bx+c ....... (1)

Let y be the altitude of the ball and x be the distance traveled by the ball.

Now by putting the points one by one, we get the required equation.

first point = (0,0)

golf ball travels a distance of 600 feet  along the ground. Therefore point = (600,0)

At altitude of 200, it travels the distance of 300 feet(mid point) = (300,200)

By putting y=0 and x=0 in (1),

we get c=0,

By putting y=0 and x=600 we get

0 = 360000a+600b ..............(2)

By putting y=200 and x=300, we get

200=90000a+300b ................(3)

Solving (2) and(3) we get, a=-0.0022 and b = 1.32

So the required equation is :

y= =0.0022+1.32x