Question 254211
The vertical height, h, in metres of a golf ball as it travels a horizontal distance, d metres, down the fairway, can be described using a quadratic function.
A pro-golfer can drive the ball 300m down the fairway before it lands. If the ball's maximum height was 15m:
:
- draw a sketch of the path of the golf ball
- find the equation of the path of the golf ball
:
Find the equation using the form ax^2 + bx = y
Coordinate at it's highest point: 150,15
a(150^2) + 150b = 15
22500a + 150 b = 15
:
Coordinate when it strikes the ground: 300,0
a(300^2) + 300b = 0
90000a = 300b = 0
:
Multiply the 1st equation by 2, subtract the 2nd equation,
45000a + 300b = 30
90000a + 300b = 0
--------------------subtraction eliminates b, find a
-45000a = 30
a = {{{30/(-45000)}}}
a = -.00067
;
Find b using the 2nd equation
90000(-.00076) + 300b = 0
-60 + 300b = 0
300b = 60
b = {{{300/60}}}
b = +.2
:
the equation of the path of the ball: h = -.00067d^2 + .2d
:
A graph of this, h = y axis; d = x axis:
{{{ graph( 300, 200, -50, 350, -10, 25, -.00067x^2+.2x) }}}
:
;
determine the two distances down the fairway when the ball is 10m above the ground.
 -.00067d^2 + .2d = 10
-.00067d^2 + .2d - 10 = 0
use the quadratic formula to find d:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this problem x = d; a = -.00067, b = .2, c = -10
{{{d = (-.2 +- sqrt(.2^2-4*-.00067*-10 ))/(2*-.00067) }}}
:
{{{d = (-.2 +- sqrt(.04 - .0268 ))/(-.00134) }}}
:
{{{d = (-.2 +- sqrt(.0132 ))/(-.00134) }}}
Two solutions
{{{d = (-.2 - .1149)/(-.00134) }}}
d = {{{(-.3149)/(-.00134)}}}
d = 235.0m
and
{{{d = (-.2 + .1149)/(-.00134) }}}
d = {{{(-.0851)/(-.00134)}}}
d = 63.5m
:
The ball will be a 10 meters at 63.5m and 235.0m