Question

An archer shoots an arrow from a height of \(1.14 \mathrm{~m}\) above ground with an initial velocity of \(47.5 \mathrm{~m} / \mathrm{s}\) and an initial 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 horizontally after approximately 2.62 seconds.

Step-by-step solution

1. Calculate the initial velocity components

We need to find the initial horizontal velocity \(v_{0x}\) and initial vertical velocity \(v_{0y}\) of the arrow. We can use the angle of release (\(35.2^{\circ}\)) to find these components using trigonometry. $$ v_{0x} = v_0 \cos(\theta)=47.5\,\text{m/s}\cdot\cos(35.2^{\circ}) $$ $$ v_{0y} = v_0 \sin(\theta)=47.5\,\text{m/s}\cdot\sin(35.2^{\circ}) $$

2. Set up the vertical motion equation

We need to find the time \((t)\) at which the arrow is flying horizontally. We know that the vertical velocity at that time will be \(0\,\text{m/s}\). We can use the equation of motion to relate the initial vertical velocity, final vertical velocity \((0\,\text{m/s})\), and the time. $$ v_{y}(t) = v_{0y} - g\cdot t $$ Where \(v_{y}(t)\) is the vertical velocity at time \(t\), \(g\) is the acceleration due to gravity \((9.8\,\text{m/s}^2)\), and \(t\) is the time in seconds.

3. Solve for time

We will now substitute \(v_{y}(t)=0\,\text{m/s}\) and the initial vertical velocity into the vertical motion equation and solve for time \(t\). $$ 0 = v_{0y} - g\cdot t\\ t=\frac{v_{0y}}{g} $$ Now, substitute the \(v_{0y}\) obtained in step 1 and \(g=9.8\,\text{m/s}^2\) to calculate the time \(t\). $$ t = \frac{47.5\,\text{m/s}\cdot \sin(35.2^{\circ})}{9.8\,\text{m/s}^2} $$

4. Calculate the time

Calculate the time \(t\) using the equation derived in step 3. $$ t \approx 2.62\,\text{s} $$ So the arrow will be flying horizontally after approximately \(2.62\,\text{seconds}\).

Key concepts

Initial Velocity Components

Understanding initial velocity components is crucial when analyzing projectile motion. Initial velocity can be broken down into two components: horizontal (\( v_{0x} \)) and vertical (\( v_{0y} \)). These components are determined by the angle at which the object is launched.

Imagine throwing a ball at an angle; it travels both forward and upward. The speed at which it moves forward (horizontal component) and the speed at which it rises (vertical component) can be found using basic trigonometry. For an object projected at an angle \(\theta\), the horizontal component is \(v_{0x} = v_0 \cos(\theta)\) and the vertical component is \(v_{0y} = v_0 \sin(\theta)\), where \(v_0\) is the initial velocity.

These components are essential for determining the overall path and behavior of the projectile, such as in the exercise where an archer shoots an arrow. The projectile will have an initial horizontal and vertical velocity based on the angle of release, which affects how long it stays in the air and how far it travels.

Equations of Motion

Equations of motion are the foundation of classical mechanics and are used to describe the motion of objects. They relate an object’s displacement, velocity, acceleration, and time.

When it comes to projectile motion, separate equations are used for the horizontal and vertical components of the motion. As there is no acceleration in the horizontal direction (ignoring air resistance), the horizontal motion can be described by the simple equation \(s_{x}=v_{0x}t\), where \(s_{x}\) is the horizontal distance traveled.

In the vertical direction, however, gravity plays a significant role, accelerating the projectile downward. The typical equations involve the initial vertical velocity (\(v_{0y}\)), the acceleration due to gravity (\(g\)), and the time (\(t\)). An example is the equation \(v_{y}(t) = v_{0y} - g\cdot t\), which computes the vertical velocity at any given time. By manipulating these equations, you can solve various parameters of motion, such as in the provided exercise, where they were used to determine when the arrow would be flying horizontally.

Acceleration Due to Gravity

Acceleration due to gravity, denoted as \(g\), is a constant force that pulls objects towards the center of the earth. On the surface of the Earth, its value is approximately \(9.8 \text{m/s}^2\), and it acts downwards towards the center of the planet.

This constant acceleration influences the motion of a projectile in the vertical direction. As highlighted in our archery example, the time it takes for an upwardly projected object to stop rising and begin falling is directly influenced by gravity. An object reaches a vertical velocity of zero when the upward force of its initial launch is completely offset by the pull of gravity.

In the exercise, the concept of gravity is pivotal, as it is used to calculate the time at which an arrow will fly horizontally. At this point, the arrow has no vertical velocity due to gravity’s acceleration, which we calculate by rearranging the vertical motion equation and solving for time. It’s a fundamental concept that helps explain not only this scenario but also an array of other phenomena, from a person jumping off a diving board to a space probe sent to orbit.