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When startled, an armadillo will leap upward. Suppose it riscs \(0.544 \mathrm{~m}\) 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) 3.70 m/s; (b) 1.74 m/s; (c) 0.155 m higher; total 0.699 m.

Step by step solution

01

Analyze Given Data

You are provided with the information: the armadillo rises to a height of \(0.544\, \text{m}\) in \(0.200\, \text{s}\). This means the displacement \(y = 0.544\, \text{m}\) and time \(t = 0.200\, \text{s}\). The acceleration due to gravity is \(g = 9.8\, \text{m/s}^2\).
02

Calculate Initial Speed using Kinematic Equation (a)

We start with the kinematic equation for upward motion: \(y = v_i t + \frac{1}{2} a t^2\). Here \(a = -9.8\, \text{m/s}^2\) (gravity acting downward), \(y = 0.544\, \text{m}\), and \(t = 0.200\, \text{s}\). Solving for \(v_i\), the initial speed, we get:\[0.544 = v_i (0.200) - \frac{1}{2} \times 9.8 \times (0.200)^2\]\[v_i = \frac{0.544 + \frac{1}{2} \times 9.8 \times (0.200)^2}{0.200}\]This requires calculation to find \(v_i\).
03

Solve for Initial Speed

Calculate the right side of the equation:\[v_i = \frac{0.544 + 0.196}{0.200}\]\[v_i = \frac{0.740}{0.200} = 3.70\, \text{m/s}\]
04

Calculate Speed at Height 0.544 m (b)

Use another kinematic equation: \(v^2 = v_i^2 + 2a (y)\). Here, \(v_i = 3.70\, \text{m/s}\), \(a = -9.8\, \text{m/s}^2\), and \(y = 0.544\, \text{m}\):\[v^2 = (3.70)^2 + 2(-9.8)(0.544)\]\[v^2 = 13.69 - 10.67\]\[v^2 = 3.02\]\[v = \sqrt{3.02}\]\[v = 1.74\, \text{m/s}\]
05

Determine Additional Height (c)

To find how much higher it goes, use \(v^2 = v_i^2 + 2a(y)\) where final speed \(v = 0\, \text{m/s}\) (at peak), initial speed \(v_i = 1.74\, \text{m/s}\) at 0.544 m height,\[0 = (1.74)^2 + 2(-9.8)y\]\[2(-9.8)y = -(1.74)^2\]\[y = \frac{-(1.74)^2}{2(-9.8)}\]Calculate to find the additional height \(y\).
06

Calculate Additional Height

Solving for \(y\):\[y = \frac{-3.03}{-19.6} = 0.155\, \text{m}\]
07

Final Calculation Step: Total Height

The total height reached from the ground is \(0.544\, \text{m} + 0.155\, \text{m} = 0.699\, \text{m}\).

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Key Concepts

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

Kinematic Equations
Kinematic equations are a foundation in understanding motion, especially in physics problems involving projectile motion. These equations help describe the motion of objects under the influence of uniform acceleration, like gravity. The kinematic equations relate different physical quantities like displacement, initial velocity, time, acceleration, and final velocity.

In our exercise, we primarily used the equation for vertical motion:
  • Displacement equation: \( y = v_i \cdot t + \frac{1}{2} a \cdot t^2 \)
This equation allows us to solve for variables like initial speed \(v_i\) or displacement \(y\) when time \(t\) and acceleration \(a\) are known.
To use these equations effectively, identify all known values and select the equation that includes the unknown variable you wish to find. Remember that for vertical motion impacted by gravity, the acceleration \(a\) is \(-9.8 \, \text{m/s}^2\), a negative value indicating direction toward Earth. These equations help unravel the components of motion by breaking down complex problems into manageable parts.
Initial Speed Calculation
Determining the initial speed of an object in projectile motion is crucial and typically involves rearranging the relevant kinematic equation. In this exercise, we used the displacement formula:
  • \( y = v_i \cdot t + \frac{1}{2} a \cdot t^2 \)
The challenge here is to solve for the initial speed \(v_i\). Given that the displacement \(y\) and time \(t\) are provided, along with the knowledge of gravitational acceleration \(a = -9.8 \, \text{m/s}^2\), we can rearrange the equation to find our unknown:
  • \( v_i = \frac{y + \frac{1}{2} a \cdot t^2}{t} \)
Here, we calculate how much effect gravity and time have on the object and solve for the initial speed. This value tells us how fast the armadillo moves upward initially, before other forces, like gravity, change its speed. Initial speed is essential because it sets the motion trajectory and determines how far or high the object will go.
Acceleration Due to Gravity
The concept of acceleration due to gravity is central to understanding projectile motion. On Earth, this constant is represented by \(g = 9.8 \, \text{m/s}^2\). It is always directed downward, towards the center of the Earth.
In problems like our armadillo exercise, this constant helps us calculate not only how displacement (or height) is affected over time, but also how velocity changes as the object moves upward or downward. For objects projected upward, gravity works against the initial velocity, slowly reducing it to zero before it reverses direction.
In every calculation involving motion against gravity:
  • Always use \(a = -9.8 \, \text{m/s}^2\), implying that gravity is decelerating upward motion.
By understanding this principle, you can apply it accurately in kinematic equations to predict the motion path, time of flight, and maximum height, all of which are essential to fully grasping the dynamics of projectile motion.

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Most popular questions from this chapter

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