91Ó°ÊÓ

Understanding how to calculate initial velocity is the first step in solving many kinematic problems. In this exercise, we are given the height the armadillo reached in a specific amount of time and need to find its initial speed as it leaves the ground.
The kinematic equation we use is:
\( y = v_0 t + \frac{1}{2} a t^2 \)
In this equation:

• \(y\) is the displacement
• \(v_0\) is the initial velocity
• \(t\) is time
• \(a\) is acceleration, which is due to gravity in this case.

We substitute the known values:
\(0.544 = v_0 (0.200) + \frac{1}{2} (-9.8)(0.200)^2\)
Solving for \(v_0\):
The equation simplifies to:
\(0.544 = 0.200 v_0 - 0.196\)
Re-ordering the terms to isolate \(v_0\):
\(0.200 v_0 = 0.740\)
Finally, we get:
\(v_0 = 3.70 \,\text{m/s}\)
This is the armadillo's initial speed as it leaves the ground. Understanding each term's role and how to rearrange the equation helps in solving similar physics problems.

Acceleration due to gravity is a constant and significant factor in kinematic equations. It affects all objects moving under Earth's gravitational pull. Typically, this value is:
\( g = 9.8 \,\text{m/s}^2\)
However, since gravity acts downwards, we often consider it as:
\( a = -9.8 \,\text{m/s}^2\)
In this exercise, gravity impacts the armadillo once it leaps into the air, slowing its upward motion until it momentarily stops at its maximum height. Understanding this helps fit the value of \(a\) correctly into the kinematic equations.
The constant value of gravitational acceleration ensures that these calculations can be generalized to various problems involving upward and downward motion.
For example, when determining the speed at a given height, we apply:
\( v = v_0 + at \)
With the previously calculated initial velocity \(v_0 = 3.70 \,\text{m/s}\)
and the time \(t = 0.200 \,\text{s}\)
we can calculate:
\( v = 3.70 + (-9.8)(0.200)\)
This simplifies to:
\( v = 1.74 \,\text{m/s}\)
This step-by-step process showcases how gravity influences the motion of objects, especially when learning to solve kinematic equations.

The maximum height an object reaches is where its upward speed is zero momentarily before it begins to fall back down. In kinematic problems, determining this height involves understanding and applying the point where velocity \(v = 0\).
Using the relevant kinematic equation:
\[ v^2 = v_0^2 + 2a y \]
Set \(v = 0\) to find the maximum height \(y\):
\[ 0 = 3.70^2 + 2(-9.8)y \]
Reordering and solving for \(y\):
\[ 3.70^2 = 19.6y \]
\[ y = \frac{3.70^2}{19.6} \]
\( y = 0.698 \,\text{m} \).
Since the armadillo already reached 0.544 meters, the additional height is:
\(0.698 - 0.544 = 0.154 \,\text{m} \)
This calculation shows how the maximum height can be segmented into its parts: the initially given height and the additional height due to the object's initial velocity and the acceleration opposing it.
Understanding this principle helps analyze the total displacement in vertical motion problems.