Problem 35 - Precalculus with Calculus Previews

Question

A famous golfer can generate a club head speed of approximately \(130 \mathrm{mi} / \mathrm{h}\) or \(v_{\mathrm{o}}=190 \mathrm{ft} / \mathrm{s}\). If the golf ball leaves the ground at an angle \(\theta_{\mathrm{o}}=\) \(45^{\circ}\), use (1) to find parametric equations for the path of the ball. What are the coordinates of the ball at \(t=2 \mathrm{~s}\) ?

Step-by-Step Solution

Answer

At \(t=2\) seconds, the ball's coordinates are \((268.70, 204.30)\) feet.

Step 1: Convert initial velocity to parametric forms

The initial velocity of the golf ball can be broken into horizontal and vertical components using the angle of projection \(\theta_{o} = 45^{\circ} \). For the horizontal component:\[v_{x} = v_{o} \cos(\theta_{o})\]\[v_{x} = 190 \cos(45^{\circ}) = 190 \cdot \frac{\sqrt{2}}{2} \approx 134.35 \text{ ft/s}\]For the vertical component:\[v_{y} = v_{o} \sin(\theta_{o})\]\[v_{y} = 190 \sin(45^{\circ}) = 190 \cdot \frac{\sqrt{2}}{2} \approx 134.35 \text{ ft/s}\]

Step 2: Determine parametric equations for position

Using these components, the parametric equations for the path of the projectile are defined by the following:The horizontal position as a function of time:\[x(t) = v_{x} \cdot t = 134.35t\]The vertical position as a function of time:\[y(t) = v_{y} \cdot t - \frac{1}{2} g t^{2} = 134.35t - \frac{1}{2} \cdot 32.2 \cdot t^2\]The value \(g = 32.2 \text{ ft/s}^2\) is the acceleration due to gravity.

Step 3: Substitute time to find coordinates at \(t=2\) seconds

Using the parametric equations derived, we substitute \(t = 2\) seconds to find the coordinates:For the horizontal position:\[x(2) = 134.35 \cdot 2 = 268.70 \text{ feet}\]For the vertical position:\[y(2) = 134.35 \cdot 2 - \frac{1}{2} \cdot 32.2 \cdot (2)^2 \]\[y(2) = 268.70 - 64.4 = 204.30 \text{ feet}\]Thus, the coordinates of the ball at \(t=2\) seconds are \((268.70, 204.30)\).

Key Concepts

Parametric Equations

In projectile motion, the path of an object can be represented using parametric equations. These equations allow us to calculate the horizontal and vertical positions of the object over time. For projectile motion, we often describe these as:

• The horizontal position: \( x(t) = v_x \cdot t \)
• The vertical position: \( y(t) = v_y \cdot t - \frac{1}{2} g t^2 \)

Here, \( v_x \) and \( v_y \) are the horizontal and vertical components of the initial velocity respectively, \( g \) is the acceleration due to gravity, and \( t \) is time. Parametric equations help us track the path of a projectile by considering time as a parameter that affects the position over both dimensions.

Initial Velocity

Initial velocity is the speed at which a projectile is launched. It is a vector quantity, meaning it has both magnitude and direction. In our example, the golf ball is hit with an initial speed of \( 190 \text{ ft/s} \). This velocity can be divided into two components:

• Horizontal component: \( v_x = v_o \cos(\theta_o) \)
• Vertical component: \( v_y = v_o \sin(\theta_o) \)

Where \( v_o \) is the initial speed and \( \theta_o \) is the angle of projection. These components allow us to use parametric equations to determine the position or trajectory of the projectile at any given time.

Angle of Projection

The angle at which an object is launched into the air is called the angle of projection. This angle plays a crucial role in determining the trajectory of a projectile. In our situation, a golf ball is hit at a \( 45^{\circ} \) angle. Such an angle is often ideal for maximizing horizontal distance under the influence of gravity.

• A higher angle might increase the vertical height but reduce the horizontal distance.
• A smaller angle may result in a lower trajectory with less vertical height but potentially greater initial horizontal distance.

By resolving the initial velocity into horizontal and vertical components based on this angle, we can better understand the motion of the projectile.

Acceleration due to Gravity

Gravity is a constant force that pulls objects toward the Earth's surface, and it greatly affects projectile motion. The standard value for the acceleration due to gravity near the Earth's surface is \( g = 32.2 \text{ ft/s}^2 \). In the context of our golf ball, this force causes the ball to decelerate as it ascends and accelerate as it falls back to the ground.

• Increases the downward velocity during ascent.
• Increases the downward velocity significantly during the descent.

This acceleration must be considered when calculating the projectile's vertical position using parametric equations, as it directly influences how high and how long the projectile stays airborne.