Question

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 cmthrough bending its limbs, compare its deceleration with that of the airman in the previous problem.

Short answer

a) -7.745 m/s

b) 1499.5 m/s²

Step-by-step solution

Given Data

• The initial velocity of the squirrel, U = 0.
• The acceleration of the squirrel,a = 10 m/s².
• The height from where the squirrel falls h = 3.0 m.

Final velocity of freely falling squirrel

a)

Final velocity can be calculated from the equation of motion as:

$V^2-U^2=2ad$

Here V is the final velocity, U is the initial velocity, a is the acceleration, and d is the distance traveled.

Substituting values in the above expression, we get,

$$\begin{gathered} (V{)}^2-(0{)}^2=2\left(10\right)\left(3\right) \\ {V}^2=60 \\ V=\sqrt{60} \\ V=7.745\text{m/s} \end{gathered}$$

Hence the velocity of the body is -7.745 m/s as the squirrel is going downward.

If bending is done in 2 cm

b)

If the squirrel is bending at 2.0 cm then, if the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with the airman in the previous problem.

The final velocity will become the initial velocity here.

In this case,

U =$-\mathbf{7}.\mathbf{745}\text{m/s}$

d =$-\mathbf{2}.\mathbf{0}\mathbf{cm}=-\text{0.02}\text{m}$

The deceleration can be calculated as:

$V^2-U^2=2ad$

Here V is the final velocity, U is the initial velocity, a is the acceleration, and d is the distance traveled.

Substituting values in the above expression, we get,

$$\begin{gathered} (0{)}^2-(-7.745{)}^2=2\times (-0.02)\times a \\ -59.98=-0.04\times a \\ a=\frac{-59.98}{-0.04} \\ a=1499.5{\text{m/s}}^{\text{2}} \end{gathered}$$

Therefore, the squirrel will be having deceleration of around 1499.5 m/s². Here negative sign represents the deceleration.