How High Does a Golf Ball Go When Driven at 24.5 Degrees?

AI Thread Summary
To determine the maximum height of a golf ball driven at an angle of 24.5 degrees, the formula H = (u^2 sin^2 θ) / (2g) is used, where u is the initial velocity and θ is the launch angle. Given that the range R is 310.24 m, the relationship H = (R/4) tan θ can also be applied. By substituting the values into these equations, the maximum height can be calculated. The discussion focuses on the physics of projectile motion and the mathematical relationships involved. Understanding these principles allows for accurate predictions of the ball's trajectory.
richardjoe05
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A golfer drives a golf ball 310.24 m down the fairway. If the ball is launched at an angle of 24.5 o to the horizontal, what is the maximum height attained by the ball during its flight?


H=1/2gt^2
 
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The maximum height attained is given by H = \frac{u^2 \sin^2 \theta}{2 g} where u is the magnitude of the initial velocity provided and θ the angle with the horizontal.

The range of the projectile is R = \frac{u^2 \sin 2 \theta}{g} which is given to be 310.24 m.

Thus we have H = \frac{R}{4} \tan \theta
 

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