# When startled an armadillo will leap upward

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

Louis Agassiz Fuertes [Public domain], via Wikimedia Commons

#### Solution:

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To do this question, we will assume:

• The air resistance to be negligible
• The armadillo leaps with constant velocity

For part (a), we will use the following equation:

$\Delta x\,=\,v_0t\,-\,\dfrac{1}{2}gt^2$

(Where $\Delta x$ is the displacement, $V_0$ is the initial velocity, $t$ is time, and $g$ is the acceleration due to gravity)

From our question we know the following:

$\Delta x\,=\,0.544$ m

$t\,=\,0.200$ s

$g\,=\,9.8\dfrac{\text{m}}{\text{s}^2}$

Let us plug these values into our equation.

$0.544=\,(0.2)(v_0)\,-\,\dfrac{1}{2}(9.81)(0.2)^2$

(solve for $v_0$)

$v_0\,=\,3.7$ m/s

For part (b), we will use the following equation:

$v\,=\,v_0\,-\,gt$

(Where $v$ is final velocity, $v_0$ is initial velocity, $g$ is the acceleration due to gravity, and $t$ is time)

From the equation we solved above, we know that $v_0\,=\,3.7$ m/s. Substituting this value into our equation gives us:

$v\,=\,3.7\,-\,(9.8)(0.2)$

$v\,=\,1.74$ m/s

For part (c), we will use the following equation:

$v^2\,=\,v_0^2-2gx$
(where $v$ is final velocity, $v_0$ is initial velocity, $g$ is the acceleration due to gravity and $x$ is displacement)

Isolate for $x$ and substitute $v_0\,=\,3.7$ m/s.

$x\,=\,\dfrac{v_0^2}{2g}$

$x\,=\,\dfrac{3.7^2}{(2)(9.8)}$

$x\,=\,0.698$ m

This value represents the total height the armadillo jumps, however, the question asks how much higher would it jump. Thus, we can figure this out by subtracting the initial height from the final height like so:

$0.698\,-\,0.544\,=\,0.154$ m.