Question

The distance, \(d,\) in metres, that a golf ball travels when struck by a golf club is given by the formula \(d=\frac{\left(v_{0}\right)^{2} \sin 2 \theta}{g},\) where \(v_{0}\) is the initial velocity of the ball, \(\theta\) is the angle between the ground and the initial path of the ball, and \(g\) is the acceleration due to gravity \(\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)\) a) What distance, in metres, does the ball travel if its initial velocity is \(21 \mathrm{m} / \mathrm{s}\) and the angle \(\theta\) is \(55^{\circ} ?\) b) Prove the identity \(\frac{\left(v_{0}\right)^{2} \sin 2 \theta}{g}=\frac{2\left(v_{0}\right)^{2}\left(1-\cos ^{2} \theta\right)}{g \tan \theta}\)

Short answer

a) The ball travels approximately 44.02 meters. b) Identity is proven using trigonometric identities and simplification.

Step-by-step solution

Identify Given Values

Values given in the problem: Initial velocity, \(v_0 = 21 \text{m/s}\), Angle, \(\theta = 55^{\text{circ}}\), and Acceleration due to gravity, \(g = 9.8 \text{m/s}^2\).

Convert Angle to Radians

The angle should be in radians for our calculations. Use the formula \( \theta_r = \frac{\pi}{180} \theta \) where \( \theta_r\) is the angle in radians.\( \theta_r = \frac{\pi}{180} \times 55^{\circ} \approx 0.9599 \text{ radians} \)

Use the Distance Formula

Plug the given values into the distance formula \( d = \frac{(v_0)^2 \sin 2 \theta}{g} \) and calculate the sine of the angle.\( \sin 2 \theta_r = \sin(2 \times 0.9599) \approx 0.9848 \)

Calculate Distance

Substitute the values into the equation to find the distance \(d\).\( d = \frac{{21^2 \times 0.9848}}{9.8} \approx 44.02 \text{ meters} \)

Prove the Identity

To prove the identity, start with the right side of the equation and transform it.Given:\[ \frac{2(v_0)^2(1- \cos^2 \theta)}{g \tan \theta} \]Recognize that \(1 - \cos^2 \theta = \sin^2 \theta \) and \(\tan \theta = \frac{\sin \theta}{\cos \theta} \).Then,\[ \frac{2(v_0)^2 \sin^2 \theta}{g \times \frac{\sin \theta}{\cos \theta}} = \frac{2(v_0)^2 \sin^2 \theta}{g \cdot \sin \theta \cdot \frac{1}{\cos \theta}} = \frac{2(v_0)^2 \sin \theta \cos \theta}{g} \]Then use the double angle identity \( \sin 2 \theta = 2 \sin \theta \cos \theta \):\[ \frac{(v_0)^2 2 \sin \theta \cos \theta}{g} = \frac{(v_0)^2 \sin 2 \theta}{g} \] proving the identity.

Key concepts

Initial Velocity

Initial velocity is the speed at which an object begins its motion. In projectile motion, it's denoted as \( v_0 \). For our golf ball problem, the initial velocity is given as 21 meters per second (m/s). This initial push is crucial as it impacts how far and fast the object will travel. The formula often seen in physics for initial velocity in projectile motion is linked to both distance and time.

Angle of Projection

The angle of projection, \( \theta \), is the angle at which the object is projected relative to the ground. For the golf ball, this angle is 55 degrees. The angle affects both the distance and the height of the projectile. It’s important because it changes the vertical and horizontal components of the initial velocity. Remember to convert degrees to radians when doing calculations: \( \theta_r = \frac{\pi}{180} \times 55^\circ \approx 0.96\) radians. This conversion is crucial for proper use of trigonometric functions in physics formulas.

Trigonometric Identities

Trigonometric identities are fundamental relationships between trigonometric functions. They come handy in simplifying complex equations. For the given problem, knowing the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \) helps in the distance formula. Also, recognizing that \( 1 - \cos^2 \theta = \sin^2 \theta \) can be useful to prove other identities. Another helpful relationship is the tangent function: \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Being comfortable with these identities can make solving projectile motion problems much simpler.

Acceleration Due to Gravity

Acceleration due to gravity, denoted as \( g \), is a constant that affects all objects near Earth's surface. Its approximate value is 9.8 meters per second squared (\( \text{m/s}^2 \)). This constant is used in the distance formula to factor in the gravitational pull. When you plug in values into the distance formula, \( d = \frac{(v_0)^2 \sin 2 \theta}{g} \), you’ll use this 9.8 m/s² to calculate how far the golf ball travels. The gravity essentially pulls the object back down towards Earth as it moves.

Distance Formula

The distance formula in projectile motion, \( d = \frac{(v_0)^2 \sin 2 \theta}{g} \), helps determine how far an object travels. Here, all components come together: the initial velocity \( v_0 \), the angle \( \theta \), and the gravity \( g \). Plug the values: initial velocity (21 m/s), angle (55 degrees converted to radians), and gravity (9.8 m/s²), then calculate \( \sin 2 \theta \). You’ll find: \( d = \frac{21^2 \times 0.9848}{9.8} \approx 44.02 \text{ meters} \). This reveals the actual distance traveled by the golf ball.