Question

An archer shoots an arrow horizontally at a target \(15 \mathrm{~m}\) away. The arrow is aimed directly at the center of the target, but it hits \(52 \mathrm{~cm}\) lower. How long did it take for the arrow to reach the target?

Short answer

The arrow took about 0.33 seconds to reach the target.

Step-by-step solution

Understand the Scenario

The arrow is shot horizontally at a target 15 meters away. The vertical displacement is 52 cm (0.52 m) downward. We need to find the time taken for the arrow to travel 15 m horizontally.

Set Up the Given Data

Horizontal distance, \(d = 15\) meters, vertical displacement, \(y = 0.52\) meters. The initial vertical velocity is \(v_{iy} = 0\) m/s because the arrow is shot horizontally.

Apply Kinematic Equation for Vertical Motion

Use the formula for vertical displacement: \(y = v_{iy}t + \frac{1}{2}gt^2\), where \(g = 9.8\, \text{m/s}^2\). Since \(v_{iy} = 0\), it simplifies to \(y = \frac{1}{2}gt^2\).

Solve for Time \(t\) using Vertical Motion Equation

Substitute \(y = 0.52\) m and \(g = 9.8\, \text{m/s}^2\) into the equation:\[0.52 = \frac{1}{2} \cdot 9.8 \cdot t^2\]\[0.52 = 4.9t^2\].Divide both sides by 4.9:\[t^2 = \frac{0.52}{4.9}\].Calculate \(t\):\[t = \sqrt{\frac{0.52}{4.9}}\].

Calculate \(t\) and Validate

Calculate the square root to find \(t\):\[t = \sqrt{\frac{0.52}{4.9}} \approx \sqrt{0.1061} \approx 0.33\, \text{s}\].Thus, it took approximately 0.33 seconds for the arrow to reach the target.

Key concepts

Horizontal Motion

In kinematics, horizontal motion refers to movement along the horizontal axis. In the context of projectile motion, like an archer shooting an arrow, the horizontal component remains unaffected by forces like gravity, assuming there's no air resistance. This means that the velocity with which the projectile is fired remains constant throughout the horizontal journey. This is what makes predicting where a projectile will land, in pure horizontal motion, simpler.

When considering the horizontal distance traveled, the formula to remember is:

• Distance = Velocity x Time

Given that our archer's arrow traveled 15 meters horizontally, it's clear that the amount of time in motion can also be deduced simply if the horizontal velocity is known.

Understanding horizontal motion allows us to see that initial conditions (like speed and angle) crucially determine the trajectory and timing.

Vertical Motion

While the arrow traveled horizontally, it was simultaneously subjected to vertical motion. Vertical motion is heavily influenced by gravitational acceleration. In our exercise, gravity pulls the arrow downwards as it travels, causing it to hit the target lower than the intended aim point.

The kinematic equation that describes this downward displacement is:

• \( y = v_{iy}t + \frac{1}{2}gt^2 \)

In our example, since the arrow was aimed horizontally, its initial vertical velocity \( v_{iy} = 0 \), simplifying the formula to:

• \( y = \frac{1}{2}gt^2 \)

Here, 'y' is the vertical displacement, 'g' stands for gravitational acceleration (9.8 m/s² localized to Earth's surface), and 't' stands for time. Calculating this gives us insight into how long the arrow was airborne.

Displacement

Displacement in physics refers to the change in position of an object. It is a vector quantity, meaning it has both magnitude and direction. In the context of the archer's arrow exercise, we are particularly interested in two types of displacement:

• Horizontal Displacement: The arrow's travel across a flat plane, which in this case is 15 meters.
• Vertical Displacement: The downward shift it experiences, which totals 0.52 meters.

Understanding displacement requires identifying these separable components and acknowledging that they contribute to the arrow's complete movement path or trajectory. By analyzing these, we can accurately determine points like impact location and flight time.

Kinematic Equations

Kinematic equations provide a framework that describes motion using minimal variables like initial velocity, final velocity, acceleration, time, and displacement. These equations help us solve motion-related problems by connecting different motion aspects in a piecewise manner.

For verticle motion, the relevant kinematic equation:

• \( y = v_{iy}t + \frac{1}{2}gt^2 \)

In situations such as the archer's arrow, where motion begins with zero vertical speed, this equation precisely calculates how far the object falls over a given time due to gravitational pull.
Utilizing the kinematic equations lets us bridge known quantities like distance and solve for unknowns, such as time. In horizontal or vertical motion, these equations simplify immensely when motion begins from rest or under consistent velocity, narrowing down the unknown variables to be solved easily.