Answer :
Thus, the golf ball lands approximately 213.12 meters from the golfer.
1. Initial velocity components:
[tex]\[ v_{0x} = 46.5 \cos 37.5^\circ \approx 36.89 \, \text{m/s} \][/tex]
[tex]\[ v_{0y} = 46.5 \sin 37.5^\circ \approx 28.31 \, \text{m/s} \][/tex]
2. Time of flight:
[tex]\[ T = \frac{2 v_{0y}}{g} = \frac{2 \times 28.31}{9.8} \approx 5.78 \, \text{s} \][/tex]
3. Range of the projectile:
[tex]\[ R = v_{0x} \cdot T = 36.89 \times 5.78 \approx 213.12 \, \text{m} \][/tex]
The golf ball lands approximately 213.03 meters from the golfer. This calculation is based on breaking down the motion into horizontal and vertical components and using projectile motion formulas. The key parameters include the initial speed, angle, and gravity.
To find the distance, we can treat this as a projectile motion problem. We will split the motion into horizontal and vertical components.
Step-by-Step Solution:
Initial speed (v0): 46.5 m/s
Angle (θ): 37.5°
First, we find the horizontal (v0x) and vertical (v0y) components of the initial velocity:
v0x = v0 * cos(θ) = 46.5 m/s * cos(37.5°) ≈ 36.87 m/s
v0y = v0 * sin(θ) = 46.5 m/s * sin(37.5°) ≈ 28.35 m/s
Using the vertical motion to determine the time the ball is in the air:
The formula for time in air (t) is given by:
t = (2 * v0y) / g
Since the ball lands back on the same level ground:
t = (2 * 28.35 m/s) / 9.8 m/s² ≈ 5.78 seconds
Now, calculate the horizontal distance (x) the ball travels, using the horizontal component of the velocity:
x = v0x * t = 36.87 m/s * 5.78 s ≈ 213.03 meters
Conclusion:
The golf ball lands approximately 213.03 meters from the golfer.