Question

A golf ball is hit with an initial angle of \(35.5^{\circ}\) with respect to the horizontal and an initial velocity of \(83.3 \mathrm{mph}\). It lands a distance of \(86.8 \mathrm{~m}\) away from where it was hit. \(\mathrm{By}\) how much did the effects of wind resistance, spin, and so forth reduce the range of the golf ball from the ideal value?

Short answer

Based on the provided information, the golf ball was hit at an initial velocity of 83.3 mph and at an angle of 35.5 degrees. The actual range where the golf ball landed is 86.8 m. Calculate the difference between the actual range of the golf ball and its ideal value (without considering air resistance, spin, etc.).

Step-by-step solution

Convert initial velocity to m/s

The initial velocity is given in mph, so we need to convert it to m/s for easier calculations: \(83.3 \mathrm{mph} \times \dfrac{1609.34 \mathrm{m}}{3600 \mathrm{s}} \approx 37.28 \mathrm{m/s}\).

Find the horizontal and vertical components of the initial velocity

Now, let's find the horizontal and vertical components of the initial velocity, using the given angle of \(35.5^{\circ}\): Horizontal component: \(V_x = V \cdot \cos(\theta) = 37.28 \cdot \cos(35.5^{\circ}) \approx 30.54 \mathrm{m/s}\) Vertical component: \(V_y = V \cdot \sin(\theta) = 37.28 \cdot \sin(35.5^{\circ}) \approx 21.33 \mathrm{m/s}\)

Calculate the time of flight for the golf ball

To calculate the time of flight for the golf ball (which is the same for the ideal case without wind resistance or spin), we use the vertical motion equation: \(t_{flight} = \dfrac{2V_y}{g} = \frac{2 \times 21.33}{9.81} \approx 4.36 \mathrm{s}\)

Calculate the ideal range of the golf ball

Now we can calculate the ideal range of the golf ball, which is the horizontal distance it would have traveled without any wind resistance or spin effects: \(R_{ideal} = V_x \times t_{flight} = 30.54 \mathrm{m/s} \times 4.36 \mathrm{s} \approx 133.15 \mathrm{m}\)

Calculate the difference between actual and ideal range

Finally, let's find the difference between the actual range of the golf ball and its ideal value: \(\Delta R = R_{ideal} - R_{actual} = 133.15 \mathrm{m} - 86.8 \mathrm{m} \approx 46.35 \mathrm{m}\) So the range of the golf ball was reduced by approximately 46.35 m due to effects like wind resistance, spin, and so forth.

Key concepts

Kinematics in Projectile Motion

Kinematics is the branch of physics that describes the motion of objects without necessarily looking at the forces that cause the motion. In the context of projectile motion, kinematics focuses on the trajectory of an object as it moves under the influence of gravity.

In projectile motion, there are two components of motion to consider: horizontal and vertical. The horizontal motion is constant because in an ideal scenario without air resistance, there is no horizontal force acting on the projectile after it's been launched. This means we can easily predict where the projectile will land horizontally by calculating the horizontal velocity multiplied by the time of flight.

However, the vertical motion is affected by gravity, which causes a projectile to accelerate downwards. We use kinematic equations to predict the vertical position and velocity at any point in time. For example, the time of flight can be calculated using the equation \( t_{flight} = \dfrac{2V_y}{g} \), where \( V_y \) is the vertical velocity, and \( g \) is the acceleration due to gravity. \( g \) is approximately \( 9.81 \text{m/s}^2 \) on Earth.

Physics of Sports and Projectile Motion

The physics of sports often involves analyzing projectile motion to improve performance and understand the effects of various factors like resistance or spin. For example, when a golf ball is hit, its trajectory is determined by the initial velocity and angle of launch. This is directly connected to the player's technique and the equipment used.

The ideal projectile motion is calculated assuming no external forces like air resistance or wind. However, in real-world scenarios, resistance, and factors such as the ball's spin, can greatly affect the ball's flight path and range. These are non-ideal conditions and are difficult to account for because of their complex and variable nature.

Therefore, understanding the basic principles of projectile motion can help athletes and coaches devise strategies to counteract these effects. For instance, in golf, players may choose particular clubs to adjust the initial velocity and angle, or they might account for wind when aiming.

Velocity Components of a Projectile

Every projectile has two main velocity components: horizontal (\( V_x \) and vertical (\( V_y \). These components are determined by the initial launch velocity and the angle of launch. The horizontal velocity affects how far the projectile will travel, while the vertical velocity affects how high it will go.

Horizontal velocity is calculated using the cosine of the launch angle: \( V_x = V \cdot \cos(\theta) \), where \( V \) is the initial velocity and \( \theta \) is the launch angle. Since there are no horizontal forces acting on the projectile in ideal conditions, this component remains constant throughout the flight.

Vertical velocity, on the other hand, is calculated using the sine of the launch angle: \( V_y = V \cdot \sin(\theta) \). Unlike horizontal velocity, vertical velocity changes over time due to the acceleration of gravity, reaching zero at the peak of the trajectory before reversing direction.