Question

An archer shoots an arrow from a height of \(1.14 \mathrm{~m}\) above ground with an initial speed of \(47.5 \mathrm{~m} / \mathrm{s}\) and a launch angle of \(35.2^{\circ}\) above the horizontal. At what time after the release of the arrow from the bow will the arrow be flying exactly horizontally?

Short answer

Answer: The arrow will be flying exactly horizontally approximately 2.76 seconds after its release from the bow.

Step-by-step solution

Break down the initial velocity into horizontal and vertical components

First, we need to find the horizontal and vertical components of the initial velocity using the given launch angle. To do this, we can use the following equations: $$ v_x = v_0 \cos{\theta}\\ v_y = v_0 \sin{\theta} $$ where \(v_x\) and \(v_y\) are the horizontal and vertical initial velocities, respectively, \(v_0\) is the initial speed, and \(\theta\) is the launch angle.

Calculate the horizontal and vertical initial velocities

Now, we can plug in the given values for \(v_0\) and \(\theta\) to find \(v_x\) and \(v_y\). $$ v_x = 47.5 \mathrm{~m} / \mathrm{s} \cos{(35.2^\circ)} \approx 38.9 \mathrm{~m} / \mathrm{s} \\ v_y = 47.5 \mathrm{~m} / \mathrm{s} \sin{(35.2^\circ)} \approx 27.0 \mathrm{~m} / \mathrm{s} $$

Find the time when the arrow is flying horizontally

We are looking for the time when the arrow's vertical velocity (\(v_y\)) is zero, i.e., \(v_y(t) = 0\). Using the equations of motion, we have: $$ v_y(t) = v_{y0} - gt $$ where \(v_{y0}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity (approximately \(9.8 \mathrm{~m} / \mathrm{s}^2\)), and \(t\) is the time. We can now solve for the time \(t\) when \(v_y(t) = 0\): $$ 0 = 27.0 \mathrm{~m} / \mathrm{s} - 9.8 \mathrm{~m} / \mathrm{s}^2 t $$ Rearranging this equation gives: $$ t = \frac{27.0 \mathrm{~m} / \mathrm{s}}{9.8 \mathrm{~m} / \mathrm{s}^2} \approx 2.76 \mathrm{~s} $$ So, the arrow will be flying exactly horizontally approximately \(2.76 \mathrm{~s}\) after its release from the bow.

Key concepts

Initial Velocity Components

Understanding the initial velocity components is crucial when analyzing projectile motion. In any projectile problem, the motion begins with an object being projected into the air at a certain angle with a specific initial speed. This initial velocity can be broken down into two components: the horizontal velocity (\(v_x\)) and the vertical velocity (\(v_y\)). These components are determined using trigonometric functions based on the launch angle.

The horizontal component, \(v_x\), is the rate at which the projectile moves along the horizontal axis. It is calculated using the cosine function of the angle: \(v_x = v_0 \cos(\theta)\), where \(v_0\) is the initial speed and \(\theta\) is the launch angle. Similarly, the vertical component, \(v_y\), reflects the rate of motion along the vertical axis and is found by \(v_y = v_0 \sin(\theta)\). Since there is no acceleration in the horizontal direction (disregarding air resistance), \(v_x\) remains constant throughout the flight. However, \(v_y\) changes due to the acceleration of gravity, affecting the object's altitude and ultimately its overall trajectory.

In our example with the archer, the initial horizontal and vertical velocities were calculated providing the foundation to solving the projectile's motion. These initial velocity components play a pivotal role in determining the projectile's behavior immediately upon release.

Equations of Motion

The equations of motion are vital for predicting the future position and velocity of a projectile at any given time. These equations incorporate acceleration, initial velocity, and time to determine the displacement and final velocity of an object in motion. For projectile motion, the force of gravity is the only constant acceleration acting on the projectile, typically represented as \(g \text{(approx. 9.8 m/s}^2)\).

For vertical motion, the primary equation to determine the vertical velocity at any time (\(t\)) is \(v_y(t) = v_{y0} - gt\), where \(v_{y0}\) represents the initial vertical velocity component. To find the time when the projectile is at a certain vertical position, we can use \(y(t) = v_{y0}t - \frac{1}{2}gt^2\), considering the initial height if necessary. This relationship is particularly useful for finding the highest point of the trajectory or when the projectile returns to its original or other specific heights.

In the archer's scenario, these equations were applied to ascertain when the arrow would be flying horizontally (i.e., when its vertical velocity equals zero). By using the appropriate formula, we can solve for \(t\) to know exactly when this event occurs.

Launch Angle

The launch angle is a defining factor in projectile motion, dictating the initial direction of the projectile's ascent and heavily influencing the shape and distance of its trajectory. Simply put, the launch angle is the angle at which an object is projected above the horizontal. It determines how much of the initial kinetic energy is distributed between vertical and horizontal motion.

A launch angle of 0 degrees results in purely horizontal motion, whereas a launch angle of 90 degrees sends the projectile straight up, resulting in no horizontal displacement. The optimal launch angle for achieving maximum range in a vacuum (without air resistance) is 45 degrees, as it equally divides the initial kinetic energy into vertical and horizontal components, allowing for maximum height and length of travel.

The launch angle affects not only the range but also the time the projectile spends in the air. In the given exercise, the launch angle of 35.2 degrees helps calculate the initial velocity components and determines the apex and time of flight. By using trigonometry with the launch angle, we isolate the initial velocity components which are essential to solving for the arrow's flight characteristics.