Math Problem Statement

A golf ball is hit with an initial velocity of 130 feet per second at an inclination of 45° to the horizontal. In physics, it is established that the height h of the golf ball is given by the function h(x) = (-32x^2 / 130^2) + x, where x is the horizontal distance the golf ball has traveled. (a) Determine the height of the golf ball after it has traveled 100 feet. (b) What is the height after it has traveled 300 feet? (c) What is h(500)? Interpret this value. (d) How far was the golf ball hit?

Solution

The function provided in the image describes the height h(x)h(x) of a golf ball in terms of its horizontal distance xx. The formula is:

h(x)=32x21302+xh(x) = \frac{-32x^2}{130^2} + x

Where:

  • xx is the horizontal distance (in feet) that the golf ball has traveled.
  • h(x)h(x) is the height (in feet) of the golf ball at that distance.

Let's go through each part of the problem:

(a) Determine the height of the golf ball after it has traveled 100 feet.

We need to calculate h(100)h(100):

h(100)=32(100)21302+100h(100) = \frac{-32(100)^2}{130^2} + 100

(b) What is the height after it has traveled 300 feet?

We need to calculate h(300)h(300):

h(300)=32(300)21302+300h(300) = \frac{-32(300)^2}{130^2} + 300

(c) What is h(500)h(500)? Interpret this value.

We need to calculate h(500)h(500) and interpret its physical meaning:

h(500)=32(500)21302+500h(500) = \frac{-32(500)^2}{130^2} + 500

(d) How far was the golf ball hit?

To find how far the golf ball was hit, we need to determine when the height h(x)h(x) is zero. This corresponds to solving the equation h(x)=0h(x) = 0:

0=32x21302+x0 = \frac{-32x^2}{130^2} + x

Let me calculate each of these values.Here are the results:

(a) Height after traveling 100 feet:

h(100)=1370016981.07feeth(100) = \frac{13700}{169} \approx 81.07 \, \text{feet}

(b) Height after traveling 300 feet:

h(300)=21900169129.59feeth(300) = \frac{21900}{169} \approx 129.59 \, \text{feet}

(c) Height after traveling 500 feet:

h(500)=450016926.63feeth(500) = \frac{4500}{169} \approx 26.63 \, \text{feet} This means that after traveling 500 feet, the ball is 26.63 feet above the ground.

(d) Distance the golf ball was hit:

The golf ball was hit a horizontal distance of: x=42258=528.125feetx = \frac{4225}{8} = 528.125 \, \text{feet} So, the ball hit the ground after traveling approximately 528.13 feet.

Let me know if you want any more details on this! Here are some related questions:

  1. How does the height change with distance after 528 feet?
  2. What is the maximum height the golf ball reaches?
  3. At what distance does the ball reach its maximum height?
  4. How would the formula change if the initial velocity was different?
  5. Can we use this equation to find the ball’s flight time?

Tip: The height function is quadratic, meaning the ball follows a parabolic trajectory!

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Functions
Projectile Motion

Formulas

Quadratic function h(x) = ax^2 + bx + c
Projectile motion formula for height

Theorems

Quadratic formula

Suitable Grade Level

Grades 9-12