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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

Expert verified
The arrow took about 0.33 seconds to reach the target.

Step by step solution

01

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.
02

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.
03

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 \(y = \frac{1}{2}gt^2\).
04

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 \(t\):\[t = \sqrt{\frac{0.52}{4.9}}\].
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A road that rises \(1 \mathrm{ft}\) for every \(100 \mathrm{ft}\) traveled horizontally is said to have a 1\% grade. Portions of the Lewiston grade, near Lewiston, Idaho, have a \(6 \%\) grade. At what angle is this road inclined above the horizontal?

A passenger walks from one side of a ferry to the other as it approaches a dock. If the passenger's velocity is \(1.50 \mathrm{~m} / \mathrm{s}\) due north relative to the ferry and \(4.50 \mathrm{~m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) west of north relative to the water, what are the direction and magnitude of the ferry's velocity relative to the water?

A crow is flying horizontally with a constant speed of \(2.70 \mathrm{~m} / \mathrm{s}\) when it releases a clam from its beak as shown in Figure 4.38. The clam lands on the rocky beach \(2.10 \mathrm{~s}\) later. Just before the clam lands, what is (a) its horizontal component of velocity and (b) its vertical component of velocity?

Two divers run horizontally off the edge of a low cliff. Diver 2 runs with twice the speed of diver 1. (a) When the divers hit the water, is the horizontal distance covered by diver 2 twice as much as, four times as much as, or equal to the horizontal distance covered by diver 1? (b) Choose the best explanation from among the following: A. The drop time is the same for both divers. B. Drop distance depends on \(t^{2}\). C. All divers in free fall cover the same distance.

A quarterback can throw a receiver a high, lazy lob pass or a low, quick bullet pass. These passes are indicated by curves 1 and 2 , respectively, in Figure 4.43. (a) The lob pass is thrown with an initial speed of \(21.5 \mathrm{~m} / \mathrm{s}\), and its time of flight is \(3.97 \mathrm{~s}\). What is its launch angle? (b) The bullet pass is thrown with a launch angle of \(25.0^{\circ}\). What is the initial speed of this pass? (c) What is the time of flight of the bullet pass?

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