# Golf

• Apr 17th 2007, 11:45 AM
imppy725
Golf
The path of the ball for many golf shots can be modelled by a quadratic function. THe path of a golf ball hit at an angle of about 10 degrees to the horizontal can be modlled by the function

h= -0.002d^2 + 0.4d

where h is the height of the ball, in meters, and d is the horizontal distance the ball travels, in meters, until it first hits the ground.

a) what is the maximum height reached by the ball?
b) what is the horizontal distance of the ball from th egolfer when the ball reaches its maximum height?
c)what is the distance does the ball travel horizontally until it first hits the ground?

if anyone can help, it would be fantastic!

Michael
• Apr 17th 2007, 12:00 PM
Jhevon
Quote:

Originally Posted by imppy725
The path of the ball for many golf shots can be modelled by a quadratic function. THe path of a golf ball hit at an angle of about 10 degrees to the horizontal can be modlled by the function

h= -0.002d^2 + 0.4d

where h is the height of the ball, in meters, and d is the horizontal distance the ball travels, in meters, until it first hits the ground.

a) what is the maximum height reached by the ball?

this is an optimization problem, find h' and set it to zero. since h can be represented by a -x^2 graph, finding where the derivative is zero, finds the maximum point.

h= -0.002d^2 + 0.4d
=> h' = -0.004d + 0.4

for max height, set h' = 0
=> -0.004d + 0.4 = 0
=> -0.004d = -0.4
=> d = 100 for max height.

so the max height occurs when the horizontal distance is 100, so the max height is:

maxh = h(100) = -0.002(100)^2 + 0.4(100)
....................= -20 + 40
....................= 20 meters

Quote:

b) what is the horizontal distance of the ball from th egolfer when the ball reaches its maximum height?
this is 100m/2 = 50m...if my memory about projectiles serves me correctly. (the distance to reach to the max height, is the distance to descend from the max height, so they will be half and half) this also makes sense since the parabola will be symmetric about its vertex

Quote:

c)what is the distance does the ball travel horizontally until it first hits the ground?
we want d when h = 0

=> 0 = -0.002d^2 + 0.4d
=> 0 = d(-0.002d + 0.4)
=> d = 0 or -0.002d + 0.4 = 0
we dont want d = 0, that's before he hit the ball

so -0.002d + 0.4 = 0
=> d = 0.4/0.002
=> d = 200 meters
• Apr 17th 2007, 12:12 PM
imppy725
what is h'?

is there another way to do it without using "diravatives" because I don't recall learning anything about diravatives. This unit is all about completing the square, finding points on a graph etc.

perhaps completing the square for part (a) would work? maybe?
• Apr 17th 2007, 12:18 PM
Soroban
Hello, Michael!

Quote:

The path of the ball for many golf shots can be modelled by a quadratic function.
The path of a golfball can be modelled by the function: .h .= .-0.002d² + 0.4d

where h is the height of the ball, in meters, and d is the horizontal distance
the ball travels, in meters, until it first hits the ground.

a) What is the maximum height reached by the ball?
b) What is the horizontal distance of the ball from the golfer
. . .when the ball reaches its maximum height?
c) What is the distance does the ball travel horizontally until it first hits the ground?

(b) The height function is a down-opening parabola.
. . Its maximum value occurs at its vertex.
The vertex of a parabola is: x = -b/2a

We have: .a = -0.002, b = 0.4
. . Hence, the vertex is at: .d = (-0.4)(-0.004) = 100

Therefore, the ball reaches maximum height when it is 100 feet from the golfer.

(a) The maximum height is: .h .= .-0.002(100²) + 0.40(100) .= .20 feet

(c) When is height equal to zero?

. . -0.002d² + 0.04d .= .0

Factor: .-0.002d(d - 200) .= .0 . . d = 0, 200

. . Of course, the ball is on the ground when d = 0.

Therefore, the ball stikes the ground 200 feet from the golfer.

• Apr 17th 2007, 12:26 PM
Jhevon
Quote:

Originally Posted by imppy725
what is h'?

is there another way to do it without using "diravatives" because I don't recall learning anything about diravatives. This unit is all about completing the square, finding points on a graph etc.

perhaps completing the square for part (a) would work? maybe?

my apologies Michael, i didn't realize you were in precalculus. Soroban has the solution for you
• Apr 17th 2007, 12:28 PM
imppy725
Quote:

Originally Posted by Jhevon
my apologies Michael, i didn't realize you were in precalculus. Soroban has the solution for you

no problem
• Apr 17th 2007, 12:29 PM
imppy725
Quote:

Originally Posted by Soroban
Hello, Michael!

(b) The height function is a down-opening parabola.
. . Its maximum value occurs at its vertex.
The vertex of a parabola is: x = -b/2a

We have: .a = -0.002, b = 0.4
. . Hence, the vertex is at: .d = (-0.4)(-0.004) = 100

Therefore, the ball reaches maximum height when it is 100 feet from the golfer.

(a) The maximum height is: .h .= .-0.002(100²) + 0.40(100) .= .20 feet

(c) When is height equal to zero?

. . -0.002d² + 0.04d .= .0

Factor: .-0.002d(d - 200) .= .0 . . d = 0, 200

. . Of course, the ball is on the ground when d = 0.

Therefore, the ball stikes the ground 200 feet from the golfer.

Hi soroban, do you mind explaining to me why you did part b before doing part a?

and will you explain each of the parts step by step for me, i have to learn why things are calculated the way they are.

thanks a lot :)
• Apr 17th 2007, 12:31 PM
Jhevon
Quote:

Originally Posted by Soroban
Hello, Michael!

(b) The height function is a down-opening parabola.
. . Its maximum value occurs at its vertex.
The vertex of a parabola is: x = -b/2a

We have: .a = -0.002, b = 0.4
. . Hence, the vertex is at: .d = (-0.4)(-0.004) = 100

Therefore, the ball reaches maximum height when it is 100 feet from the golfer.

I'm not sure about this one Soroban. remember, d is not the distance from the golfer per say, it is the distance until the ball hits the ground. so it will be the total distance travelled if the ball reached the maximum height, the the ball must have reached the maximum height before 100 meters. i claim it is half-way, that is at 50 meters away from the golfer
• Apr 17th 2007, 12:34 PM
Jhevon
Quote:

Originally Posted by imppy725
Hi soroban, do you mind explaining to me why you did part b before doing part a?

and will you explain each of the parts step by step for me, i have to learn why things are calculated the way they are.

thanks a lot :)

Soroban feels as i do that the questions were asked in the wrong order, since to answer (a), you would have to calculate the answer to (b), so he did that first.

which steps are you confused about? the ones for (a), (b), (c) or all?
• Apr 17th 2007, 12:38 PM
imppy725
Quote:

Originally Posted by Jhevon
Soroban feels as i do that the questions were asked in the wrong order, since to answer (a), you would have to calculate the answer to (b), so he did that first.

which steps are you confused about? the ones for (a), (b), (c) or all?

pretty much all.
• Apr 17th 2007, 01:02 PM
Jhevon
Quote:

Originally Posted by imppy725
The path of the ball for many golf shots can be modelled by a quadratic function. THe path of a golf ball hit at an angle of about 10 degrees to the horizontal can be modlled by the function

h= -0.002d^2 + 0.4d

where h is the height of the ball, in meters, and d is the horizontal distance the ball travels, in meters, until it first hits the ground.

Ok, you have the calculations, i'll assume you are comfortable with them (unless you tell me otherwise). now i'll explain the logic behind the calculations.

Quote:

a) what is the maximum height reached by the ball?
see the diagram below. we see that the graph of h will be a downward opening parabola, since it will be a -x^2 graph. the maximum height then, is the highest point on the graph, since this poit will have the largest y-value for a corresponding x-value if we graph it.

this point is the resting point of the parabola and is also called the "vertex" in precalculus, the formula to find the x-coordinate for the vertex is:

x = -b/2a

what does a and b mean? well they are just the coefficients in an x^2 graph of the following form:

y = ax^2 + bx + c

so a is the coefficient of x^2, b is the coefficient of x, and c is the lone constant.

in your graph, we have h instead of y and d instead of x, so yours in in the form:

h = ad^2 + bd + c

so a = -0.002, b = 0.4 and c = 0

doing that, we get a value for d where the vertex occurs (which is what question (b) is asking for, that's why Soroban did it first), and so, to find h, we plug the value for d we found into the equation. this gives the maximum height.

Quote:

b) what is the horizontal distance of the ball from th egolfer when the ball reaches its maximum height?
so you see how we got d from the part above. now there is a trick to this question. d is not the distance from the golfer, it is the total distance travelled after the ball lands, see the diagram. so d is the distance from one foot of the parabola to the next, a parabola is symmetric about it's vertex, so the distance travelled to reach the max height, is half the total distance travelled. so after finding d, we divide it by 2

Quote:

c)what is the distance does the ball travel horizontally until it first hits the ground?
the ball hits the ground when the height is zero, so we simply plug in h = 0 into our equation and solve for d. we get 2 values, one is trivial and the other is the answer

got it?