Consider a grey squirrel falling out of a tree to the ground.

(a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel’s velocity just before hitting the ground, assuming it fell from a height of 3.0 m.

(b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.

Solution:

Part A

We are given the following: v_0=0 \ \text{m/s}; a=-9.80 \ \text{m/s}^2; and \Delta x=-3.0\ \text{m}.

Note that the acceleration is due to gravity and its value is constant at a=-9.80 \ \text{m/s}^2. Also, the distance x, is negative because of the direction of motion. For free-fall, downward motion is considered negative. To solve for the velocity just before it hits the ground, we will solve v_fin the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Substituting the given values:

\begin{align*}
\left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\
v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\
v_f & = \sqrt{\left(0\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(-3.0\:\text{m}\right)} \\
v_f & = 7.7\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We are given the following: v_f=0 \ \text{m/s}; v_0=7.7 \ \text{m/s}; and \Delta x=0.02 \ \text{m}

To solve for the acceleration we shall use the same formula as that in Part A. Solving for the acceleration a:

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values:

\begin{align*}
a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\
a & =\frac{\left(0\:\text{m/s}\right)^2-\left(7.7\:\text{m/s}\right)^2}{2\left(0.02\:\text{m}\right)} \\
a & =-1.5\times 10^3\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

This is approximately 3 times the deceleration of the airman from the previous problem, who was falling from thousands of meters high.