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?

When startled an armadillo will leap upward

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.

 

Final Answers:

a) 3.7 m/s

b) 1.74 m/s

c) 0.154 m

 

This question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 44.

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