Math Problem Statement

The hole on a level, elevated golf green is a horizontal distance of 120 meters from the tee and a vertical distance of 12.8 meters above the tee. A golfer hits a ball at an angle of 9.2 degrees and makes a hole in one. What was the initial speed of the ball?

Solution

We can solve this problem by analyzing the horizontal and vertical components of the ball's motion using kinematic equations. Let's break it down:

Given:

  • Horizontal distance to the hole, dx=120md_x = 120 \, \text{m}
  • Vertical height difference, h=12.8mh = 12.8 \, \text{m}
  • Launch angle, θ=9.2\theta = 9.2^\circ
  • Acceleration due to gravity, g=9.8m/s2g = 9.8 \, \text{m/s}^2
  • The ball lands in the hole, which means its final position matches the target.

We need to find the initial speed of the ball, v0v_0.

Step 1: Horizontal motion equation

The horizontal distance covered is given by: dx=v0xtd_x = v_{0x} \cdot t where v0x=v0cos(θ)v_{0x} = v_0 \cdot \cos(\theta) is the horizontal component of the initial velocity, and tt is the time of flight.

t=dxv0cos(θ)t = \frac{d_x}{v_0 \cdot \cos(\theta)}

Step 2: Vertical motion equation

For vertical motion, we use the equation: y=v0yt12gt2y = v_{0y} \cdot t - \frac{1}{2} g t^2 where y=hy = h, and v0y=v0sin(θ)v_{0y} = v_0 \cdot \sin(\theta) is the vertical component of the initial velocity.

Substituting the expression for tt from the horizontal motion equation into the vertical motion equation: h=v0sin(θ)dxv0cos(θ)12g(dxv0cos(θ))2h = v_0 \cdot \sin(\theta) \cdot \frac{d_x}{v_0 \cdot \cos(\theta)} - \frac{1}{2} g \left( \frac{d_x}{v_0 \cdot \cos(\theta)} \right)^2

Simplifying the equation: h=dxtan(θ)gdx22v02cos2(θ)h = d_x \cdot \tan(\theta) - \frac{g d_x^2}{2 v_0^2 \cdot \cos^2(\theta)}

Step 3: Solve for v0v_0

Now we can solve for v0v_0. Rearranging the above equation:

h=dxtan(θ)gdx22v02cos2(θ)h = d_x \cdot \tan(\theta) - \frac{g d_x^2}{2 v_0^2 \cdot \cos^2(\theta)} gdx22v02cos2(θ)=dxtan(θ)h\frac{g d_x^2}{2 v_0^2 \cdot \cos^2(\theta)} = d_x \cdot \tan(\theta) - h v02=gdx22cos2(θ)(dxtan(θ)h)v_0^2 = \frac{g d_x^2}{2 \cdot \cos^2(\theta) \cdot (d_x \cdot \tan(\theta) - h)}

Taking the square root: v0=gdx22cos2(θ)(dxtan(θ)h)v_0 = \sqrt{\frac{g d_x^2}{2 \cdot \cos^2(\theta) \cdot (d_x \cdot \tan(\theta) - h)}}

Now, let's plug in the values:

  • g=9.8m/s2g = 9.8 \, \text{m/s}^2
  • dx=120md_x = 120 \, \text{m}
  • θ=9.2\theta = 9.2^\circ (convert to radians: θ=9.2×π180\theta = 9.2 \times \frac{\pi}{180})
  • h=12.8mh = 12.8 \, \text{m}

I'll calculate the value of v0v_0.The initial speed of the ball was approximately 104.46m/s104.46 \, \text{m/s}.

Would you like a detailed explanation of any part of the solution?

Here are five related questions that could deepen your understanding:

  1. How do the horizontal and vertical components of velocity affect the trajectory?
  2. What happens to the initial speed if the launch angle increases?
  3. How would air resistance alter the calculations for a golf ball?
  4. What would be the effect of changing the elevation of the golf hole?
  5. Can you derive the formula for the time of flight in projectile motion?

Tip: In projectile motion problems, always break the motion into horizontal and vertical components for easier analysis.

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Math Problem Analysis

Mathematical Concepts

Projectile Motion
Kinematics
Trigonometry

Formulas

d_x = v_{0x} * t
v_{0x} = v_0 * cos(θ)
y = v_{0y} * t - (1/2) * g * t^2
v_{0y} = v_0 * sin(θ)
v_0 = sqrt(g * d_x^2 / (2 * cos^2(θ) * (d_x * tan(θ) - h)))

Theorems

Kinematic Equations of Motion
Trigonometric Functions in Motion

Suitable Grade Level

Undergraduate Physics