The tee of the world’s longest par 3 sits atop South Africa’s Hanglip Mountain at 400 m above the green and can only be reached by helicopter.  The horizontal distance to the green is 359 m.   Let g = 9.8 m/s2.  Neglect air resistance and answer the following questions.
If a golfer launches a shot that is 28.3 degrees with respect to the horizontal, what initial velocity must she give the ball so that it just reaches the green and how much time does it take the ball to reach the green?

Solve for this time.

We are tasked with finding the initial velocity and the time it takes for the ball to reach the green when hit from the top of a mountain with a launch angle of 28.3° from the horizontal. Here's how we can solve it step-by-step:

### Given dataWe are tasked with finding the initial velocity and the time it takes for the ball to reach the green when hit from the top of a mountain with a launch angle of 28.3° from the horizontal. Here's how we can solve it step-by-step:

### Given 
- Height of the mountain, \( h = 400 \, \text{m} \)
- Horizontal distance to the green, \( d = 359 \, \text{m} \)
- Launch angle, \( \theta = 28.3^\circ \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

We need to find two things:
1. The initial velocity \( v_0 \)
2. The time of flight \( t \)

### Step 1: Split the motion into horizontal and vertical components
The horizontal and vertical motions can be treated separately using the following kinematic equations:

- **Horizontal motion** (constant velocity):
  \[
  x = v_{0x} \cdot t
  \]
  where \( v_{0x} = v_0 \cos \theta \), and \( x = d \).

- **Vertical motion** (accelerated due to gravity):
  \[
  y = v_{0y} \cdot t - \frac{1}{2} g t^2
  \]
  where \( v_{0y} = v_0 \sin \theta \), and \( y = -400 \, \text{m} \) (since the ball is falling 400 m down to the green).

### Step 2: Horizontal motion equation
From the horizontal motion equation:
\[
x = v_{0x} \cdot t
\]
\[
359 = (v_0 \cos 28.3^\circ) \cdot t
\]
This gives us the first equation:
\[
t = \frac{359}{v_0 \cos 28.3^\circ}
\]
We will substitute this expression for \( t \) into the vertical motion equation.

### Step 3: Vertical motion equation
For the vertical motion, we use the equation:
\[
y = v_{0y} \cdot t - \frac{1}{2} g t^2
\]
Substitute \( y = -400 \, \text{m} \) and \( v_{0y} = v_0 \sin 28.3^\circ \):
\[
-400 = (v_0 \sin 28.3^\circ) \cdot t - \frac{1}{2} (9.8) t^2
\]
Now substitute the expression for \( t \) from the horizontal equation:
\[
t = \frac{359}{v_0 \cos 28.3^\circ}
\]
into the vertical equation to solve for \( v_0 \).

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**Tip:** In projectile problems, breaking down the motion into horizontal and vertical components is key to understanding the trajectory.

The tee of the world’s longest par 3 sits atop South Africa’s Hanglip Mountain at 400 m above the green and can only be reached by helicopter. The horizontal distance to the green is 359 m. Let g = 9.8 m/s². Neglect air resistance and answer the following questions. If a golfer launches a shot that is 28.3 degrees with respect to the horizontal, what initial velocity must she give the ball so that it just reaches the green and how much time does it take the ball to reach the green?

Solve a projectile motion problem where a golfer launches a ball from 400m above the green with an angle of 28.3 degrees. Learn how to calculate the initial velocity and time of flight using kinematic equations. Suitable for high school physics students.

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