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Happy Gilmore hits a golf ball off a tee at an angle of 25 degrees and a velocity of 98

m/s. What are the horizontal and vertical velocities of the ball as it comes off the club?
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Final answer:

Happy Gilmore's golf ball has a horizontal velocity of approximately 88.82 m/s and a vertical velocity of approximately 41.41 m/s, calculated using trigonometric functions based on the given initial velocity and launch angle.

Step-by-step explanation:

To determine the horizontal and vertical velocities of the golf ball hit by Happy Gilmore, we can use the following trigonometric relationships:

The horizontal velocity (Vx) is Vx = V0 × cos(θ), where V0 is the initial velocity and θ is the launch angle.

The vertical velocity (Vy) is Vy = V0 × sin(θ).

Given that the initial velocity (V0) is 98 m/s and the angle of launch (θ) is 25 degrees, we can calculate:

Horizontal velocity (Vx):

Vx = 98 m/s × cos(25°) = 98 m/s × 0.9063 = 88.82 m/s (approx).

Vertical velocity (Vy):

Vy = 98 m/s × sin(25°) = 98 m/s × 0.4226 = 41.41 m/s (approx).

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