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When startled, an armadillo will leap upward. Suppose it rises \(0.544 \mathrm{~m}\) in the first \(0.200 \mathrm{~s}\). (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of \(0.544 \mathrm{~m}\) ? (c) How much higher does it go?

Short Answer

Expert verified
(a) Initial speed is \(2.62 \text{ m/s}\). (b) Speed at \(0.544 \text{ m}\) is \(0 \text{ m/s}\). (c) Total height is \(0.894 \text{ m}\).

Step by step solution

01

Identify Known Values

We are given the height (displacement) the armadillo reaches, \( \Delta y = 0.544 \text{ m} \), and the time taken to reach this height, \( t = 0.200 \text{ s} \). The acceleration due to gravity is \( g = 9.81 \text{ m/s}^2 \).
02

Determine Initial Velocity (Part a)

Using the kinematic equation: \( \Delta y = v_i t + \frac{1}{2} a t^2 \), where \( a = -g \) (as the acceleration due to gravity is downward). Rearrange the equation for \( v_i \) to get: \[ v_i = \frac{\Delta y - \frac{1}{2} a t^2}{t} \]. Substitute the known values: \[ v_i = \frac{0.544 \text{ m} - \frac{1}{2}(9.81 \text{ m/s}^2)(0.200 \text{ s})^2}{0.200 \text{ s}} \]. Simplifying gives \( v_i \approx 2.62 \text{ m/s} \).
03

Calculate Final Speed at 0.544 m (Part b)

Use the kinematic equation: \( v_f^2 = v_i^2 + 2a\Delta y \) to find the speed at \( \Delta y = 0.544 \text{ m} \). First, we need to solve for the new velocity \((v_f)\), knowing initial velocity \(v_i\) and displacement. \[ v_f = \sqrt{v_i^2 + 2(-g)(0.544 \text{ m})} \]. Substitute \( v_i \approx 2.62 \text{ m/s} \) calculated earlier: \[ v_f = \sqrt{(2.62 \text{ m/s})^2 - 2(9.81 \text{ m/s}^2)(0.544 \text{ m})} \]. This results in \( v_f = 0 \text{ m/s} \).
04

Determine Maximum Height (Part c)

To find how much higher the armadillo goes, use \( v_f = 0 \text{ m/s} \), indicating the highest point. Use the kinematic formula: \( v_f^2 = v_i^2 + 2a\Delta h \), where \( v_f = 0 \text{ m/s} \). Rearrange for \( \Delta h \): \[ 0 = v_i^2 - 2g\Delta h \]. Solve for \( \Delta h \): \[ \Delta h = \frac{v_i^2}{2g} \]. Substituting \( v_i \approx 2.62 \text{ m/s} \): \[ \Delta h = \frac{(2.62 \text{ m/s})^2}{2 (9.81 \text{ m/s}^2)} \]. \( \Delta h \approx 0.35 \text{ m} \).
05

Calculate Total Maximum Height

Add the additional height \( \Delta h \approx 0.35 \text{ m} \) to the height \( 0.544 \text{ m} \) initially reached. Total height the armadillo reaches is \( 0.544 \text{ m} + 0.35 \text{ m} = 0.894 \text{ m} \).

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Key Concepts

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

Initial Velocity
In kinematics, **initial velocity** (\( v_i \) ) is the speed at which an object begins its motion. It's crucial for predicting future motion. When analyzing an armadillo's leap, we use this concept to calculate how fast it started its journey upwards from the ground.To find the initial velocity, we consider the distance the armadillo rises in the given time and the impact of gravity pulling it back down. The kinematic equation that helps us calculate this is:- \( \Delta y = v_i t + \frac{1}{2} a t^2 \)Here:- \( \Delta y \) is the change in vertical position (0.544 m in our problem)- \( t \) is the time taken (0.200 s)- \( a \) is the acceleration (which is negative due to gravity)By rearranging the equation for \( v_i \), we solve for:- \( v_i = \frac{\Delta y - \frac{1}{2} a t^2}{t} \)Calculations show an initial speed of approximately 2.62 m/s. This means the armadillo left the ground with this velocity, propelling itself upwards against the force of gravity.
Acceleration Due to Gravity
The **acceleration due to gravity** is a fundamental concept in physics, denoted by \( g \) and having a standard value of 9.81 m/s². It represents the rate at which an object accelerates downwards due to Earth's gravitational pull.When an object is moving straight up, like the armadillo, this acceleration works against it, slowing it down until it comes to a temporary stop at its peak. In our calculations, we consider gravity as a negative value because it's acting in the opposite direction of the initial motion.Gravity plays into many forms of motion:- It causes projectiles to curve downward.- It affects the time objects spend in the air.- It reduces the vertical speed of objects moving upwards.Using gravity in calculations helps predict when and where an object will reach its maximum height and return to the ground. This concept is pivotal when computing changes in velocity and distance traveled during an object's flight.
Maximum Height
Understanding **maximum height** is key to studying projectile motion. It's where the armadillo stops moving upward and starts descending. At this point, its velocity is zero for a moment.The maximum height is calculated by determining how much higher an object goes after a given point. In our scenario, the armadillo first reaches a height of 0.544 m, but because it has momentum, it continues to rise even further. We use the kinematic equation to find the additional height \( \Delta h \) beyond this initial height:- \( \Delta h = \frac{v_i^2}{2g} \)Here:- \( v_i \) is the initial velocity calculated earlier- \( g \) is acceleration due to gravityFor the armadillo, \( \Delta h \) comes out to be about 0.35 m.The total maximum height combines the armadillo's initial rise and the extra height, totaling approximately 0.894 m. This shows how an initial push translates to significant vertical travel.
Final Velocity
**Final velocity** is the speed of an object at the end of its motion. When the armadillo reaches the peak of its jump, its final velocity is 0 m/s because it temporarily stops before falling back down.To find the final velocity when the armadillo reaches a specific height like 0.544 m, we use another kinematic formula:- \( v_f^2 = v_i^2 + 2a\Delta y \)At this height, the final velocity calculation confirms our intuition since the armadillo just reaches the top of its arc without excess speed.For sub-scenarios where final velocity isn’t 0, this calculation helps determine:- The actual speed upon hitting a certain height or making contact with the ground.- The influence of initial speed and gravitational pull on motion end states.Understanding final velocity aids in analyzing how forces interact over time and understanding real-world applications like predicting landing speeds or the impact forces involved.

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