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 cm through bending its limbs, compared its deceleration with that of the airman in the previous problem.

Answer 1

Answer:

(a). v = 7.745 m/s

(b). a = -1500 m²/s

Explanation:

In the question,

(a).

We know that the height of the squirrel from the ground, h = 3 m

Now,

From the equation of the motion we can say that,

$s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\3=0+5t^{2}\\t^{2}=\frac{3}{5}\\t^{2}=0.6\\t=0.77\,s$

Also,

$v^{2}-u^{2}=2as\\v^{2}=2(10)(3)=60\\v=7.745\,m/s$

Therefore, the velocity of squirrel before hitting the ground is 7.745 m/s.

(b).

The squirrel stops in the distance of 2 cm after falling through the height.

So,

The deceleration of the squirrel is = a

So, using the law of the motion in the last 2 cm of distance travelled.

Where,

u = 7.745 m/s

So,

$v^{2}-u^{2}=2as\\0-(7.745)^{2}=2a(\frac{2}{100})\\-60=0.04a\\a=-1500\,m^{2}/s$

Therefore, the deceleration of the squirrel is given by, a = -1500 m²/s