Question

A golfer launches a tee shot down a horizontal fairway and it follows a path given by \(\mathbf{r}(t)=\left\langle a t,(75-0.1 a) t,-5 t^{2}+80 t\right\rangle,\) where \(t \geq 0\) measures time in seconds and \(\mathbf{r}\) has units of feet. The \(y\) -axis points straight down the fairway and the z-axis points vertically upward. The parameter \(a\) is the slice factor that determines how much the shot deviates from a straight path down the fairway. a. With no slice \((a=0),\) sketch and describe the shot. How far does the ball travel horizontally (the distance between the point the ball leaves the ground and the point where it first strikes the ground)? b. With a slice \((a=0.2),\) sketch and describe the shot. How far does the ball travel horizontally? c. How far does the ball travel horizontally with \(a=2.5 ?\)

Short answer

Answer: When the slice factor is \(a=2.5\), the horizontal distance traveled by the golf ball along the y-axis is 1160 feet.

Step-by-step solution

Find when the ball first strikes the ground (\(z=0\))

To find when the ball first strikes the ground, we need to set the z-component of the function \(\mathbf{r}(t)\) to 0: $$-5t^2 + 80t = 0$$ Now, we need to solve this quadratic equation for t.

Solve the quadratic equation

To solve the quadratic equation, we factor out t: $$t(-5t + 80) = 0$$ This gives us two possible values for t: \(t = 0\) or \(t=\frac{80}{5}=16\). Since t = 0 corresponds to the starting point when the ball is launched, the time when the ball first strikes the ground is \(t=16\) seconds.

Part a: No slice (a=0)

Now we plug \(a=0\) and \(t=16\) into the y-component of \(\mathbf{r}(t)\) to find the horizontal distance: $$y(t)= (75 - 0.1 \cdot 0)t = 75t$$ $$y(16) = 75\cdot16 = 1200\,\text{ft}$$ Thus, with no slice, the golf ball travels horizontally 1200 feet down the fairway. The shot would be a high and straight shot with a parabolic trajectory.

Part b: Slice (a=0.2)

Plug \(a=0.2\) and \(t=16\) into the y-component and x-component of \(\mathbf{r}(t)\) to find the horizontal distance and slice amount: $$x(t) = 0.2t$$ $$y(t) = (75 - 0.1 \cdot 0.2)t = 74.98t$$ $$x(16) = 0.2\cdot 16 = 3.2\,\text{ft}$$ $$y(16) = 74.98\cdot 16 = 1199.68\,\text{ft}$$ With a slice factor of 0.2, the golf ball moves 3.2 feet to the right (for a right-handed golfer) and 1199.68 feet down the fairway. The trajectory would be similar to the shot with no slice, but it will have a slight curve to the right.

Part c: Horizontal distance with a=2.5

Plug \(a=2.5\) and \(t=16\) into the y-component of \(\mathbf{r}(t)\) to find the horizontal distance: $$y(t) = (75 - 0.1 \cdot 2.5)t = 72.5t$$ $$y(16) = 72.5\cdot 16 = 1160\,\text{ft}$$ With a slice factor of 2.5, the golf ball travels horizontally 1160 feet down the fairway. The shot's trajectory would be strongly curved to the right, and it would not travel as far down the fairway as the shots with lower slice factors.

Key concepts

Trajectory Calculation

Calculating the trajectory of an object, such as a golf ball, launched into the air involves determining how the object moves through space. Each parameter in the parametric equation influences the object's path. The trajectory is essentially the path followed by the object due to forces acting upon it, such as gravity in this case.

In the given problem, the parametric equation \( \mathbf{r}(t) = \langle at, (75-0.1a)t, -5t^2+80t \rangle \) is used to describe the ball's position in three dimensions. The time \( t \) allows us to track this position. The equation tells us how far the ball moves down the fairway and how high it ascends.

The "slice factor" \( a \) affects how much the ball veers from a straight path down the fairway, showing the importance of considering all parametric components when predicting trajectories. For instance, when \( a = 0 \), the path is straight, but as \( a \) increases, the ball's path curves more noticeably, reflecting realistic golf shots where direction can deviate due to spin or mishits.

When finding out where the ball first strikes the ground, you calculate when the vertical component (represented by the \( z \)-component in the equation) equals zero. This represents the time \( t \) when the ball "lands." This is a common approach in physics or engineering to understand distances and trajectories.

Three-Dimensional Vectors

Vectors in three dimensions are a powerful tool for understanding and calculating positions in space. In our exercise, the position vector \( \mathbf{r}(t) \) illustrates how the ball moves in three directions, represented by \( x, y, \) and \( z \) components.

• The \( x \)-component (\( at \)) represents the lateral movement or slice, measuring how far the ball is off the direct line down the fairway.
• The \( y \)-component ((\( 75-0.1a)t \)) tracks how far along the fairway the ball travels.
• The \( z \)-component (-\( 5t^2 + 80t \)) shows the ball's vertical motion, illustrating how high it goes before gravity pulls it down. It is a function of time and forms a parabolic shape.

Such vectors offer a complete picture of motion by focusing on direction and magnitude — essential for trajectory calculations. The ability to visualize and compute in three dimensions is vital in fields ranging from physics to computer graphics, aiding in problem-solving and simulation.

Quadratic Equations

Quadratic equations often appear in physics and engineering problems when dealing with motions that involve acceleration, like the trajectory of a golf ball under gravity. They take the form \( ax^2 + bx + c = 0 \), and solutions can be found using factoring, the quadratic formula, or completing the square.

In our context, we see the vertical component of the trajectory, \( -5t^2 + 80t = 0 \), fitting this model. Factoring gives \( t(-5t + 80) = 0 \), which simplifies to \( t = 0 \) or \( t = 16 \). Here, \( t = 0 \) is the launch, and \( t = 16 \) seconds is when the ball hits the ground.

• Factoring is a straightforward method to find the roots of quadratic equations when they can easily be broken into simpler binomials.
• Understanding the roots of the quadratic helps predict when events occur, such as when the ball returns to the ground.

Quadratics encapsulate much of the dynamic world around us, illustrating how parabolic motions appear in contexts such as ballistics or even economics, making it a critical concept across multiple domains.