**Physics Problem: Projectile Motion****Scenario:**A roadrunner is cruising at a velocity of 11 m/s without realizing that the road beneath its feet is about to end at the edge of a cliff, which is approximately 30 meters above a flat valley below. Fortunately, the roadrunner can fly and successfully escaped.**Question:**If the roadrunner could not fly and was unaffected by air resistance, how far horizontally from the base of the cliff would it land in the valley below?**Graphical Explanation:**The problem involves calculating the horizontal distance a projectile (in this case, the roadrunner) would travel when it falls from a given height (30 meters) with an initial horizontal velocity (11 m/s).To solve this, we can use the kinematic equations of motion.1. **Vertical Motion Calculation:** - The initial vertical velocity (\( v_{iy} \)) is 0 m/s since the roadrunner runs horizontally. - The height (h) is 30 meters. - The acceleration due to gravity (g) is approximately 9.8 m/s². Using the equation for the vertical motion: \[ h = \frac{1}{2} gt^2 \] \[ 30 = \frac{1}{2} (9.8) t^2 \] \[ 30 = 4.9 t^2 \] \[ t^2 = \frac{30}{4.9} \] \[ t^2 \approx 6.122 \] \[ t \approx 2.47 \text{ seconds} \]2. **Horizontal Motion Calculation:** - Given that the horizontal velocity (\( v_{ix} \)) is 11 m/s, and the time of flight (t) is 2.47 seconds. - The horizontal distance (d) is given by: \[ d = v_{ix} \times t \] \[ d = 11 \times 2.47 \] \[ d \approx 27.17 \text{ meters} \]Therefore, without the ability to fly, the roadrunner would hit the valley floor approximately 27.17 meters horizontally from the edge of

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Description: **Physics Problem: Projectile Motion****Scenario:**A roadrunner is cruising at a velocity of 11 m/s without realizing that the road beneath its feet is about to end at the edge of a cliff, which is approximately 30 meters above a flat valley below.

The roadrunner is running at a speed of 11 m/s on a road that is 30 m above a valley.

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A roadrunner is cruising at 11 m/s without knowing the road under its feet is about to run out, at the edge of a cliff, some 30 m above a flat valley below. Fortunately a roadrunner CAN fly so it just flew away. If indeed the roadrunner couldn't fly and would not be affected by air drag, how far horizontally from the cliff will it hit the valley (flat) below?

A roadrunner is cruising at 11 m/s without knowing the road under its feet is about to run out, at the edge of a cliff, some 30 m above a flat valley below. Fortunately a roadrunner CAN fly so it just flew away. If indeed the roadrunner couldn't fly and would not be affected by air drag, how far horizontally from the cliff will it hit the valley (flat) below?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Displacement, Velocity and Acceleration

In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.

Linear Displacement

The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.

Topic Video

Question

**Physics Problem: Projectile Motion**

**Scenario:**
A roadrunner is cruising at a velocity of 11 m/s without realizing that the road beneath its feet is about to end at the edge of a cliff, which is approximately 30 meters above a flat valley below. Fortunately, the roadrunner can fly and successfully escaped.

**Question:**
If the roadrunner could not fly and was unaffected by air resistance, how far horizontally from the base of the cliff would it land in the valley below?

**Graphical Explanation:**
The problem involves calculating the horizontal distance a projectile (in this case, the roadrunner) would travel when it falls from a given height (30 meters) with an initial horizontal velocity (11 m/s).

To solve this, we can use the kinematic equations of motion.

1. **Vertical Motion Calculation:**
- The initial vertical velocity (\( v_{iy} \)) is 0 m/s since the roadrunner runs horizontally.
- The height (h) is 30 meters.
- The acceleration due to gravity (g) is approximately 9.8 m/s².

Using the equation for the vertical motion:
\[
h = \frac{1}{2} gt^2
\]
\[
30 = \frac{1}{2} (9.8) t^2
\]
\[
30 = 4.9 t^2
\]
\[
t^2 = \frac{30}{4.9}
\]
\[
t^2 \approx 6.122
\]
\[
t \approx 2.47 \text{ seconds}
\]

2. **Horizontal Motion Calculation:**
- Given that the horizontal velocity (\( v_{ix} \)) is 11 m/s, and the time of flight (t) is 2.47 seconds.
- The horizontal distance (d) is given by:
\[
d = v_{ix} \times t
\]
\[
d = 11 \times 2.47
\]
\[
d \approx 27.17 \text{ meters}
\]

Therefore, without the ability to fly, the roadrunner would hit the valley floor approximately 27.17 meters horizontally from the edge of

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