A 580-g squirrel with a surface area of 855 cm2 falls from a 5.2-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the cross-sectional area of the squirrel can be approximated as a rectangle of width 11.1 cm and length 22.2 cm. Note, the squirrel may not reach terminal velocity by the time it hits the ground. Give the squirrel's terminal velocity, not it's velocity as it hits the ground.

I have gotten the correct terminal velocity for this portion of the question but do not know how to find the answer to the following part:
What will be the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance?

vf^2=2*9.8*height

To find the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance, you can use the principle of conservation of energy.

First, let's assume that the person falls from a height of h meters. The potential energy (PE) the person has at the top is given by:

PE = m * g * h

where m is the mass of the person (58.0 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.

Next, let's assume that the person reaches a final velocity of v m/s just before hitting the ground. The kinetic energy (KE) the person has just before impact is given by:

KE = (1/2) * m * v^2

According to the principle of conservation of energy, the potential energy at the top is equal to the kinetic energy just before impact. Therefore:

PE = KE
m * g * h = (1/2) * m * v^2

Simplifying and solving for v, we get:

v^2 = 2 * g * h
v = sqrt(2 * g * h)

Given that g is approximately 9.8 m/s^2, and assuming that the height of the fall is negligible compared to the squirrel's fall from the tree, we can use the same velocity you calculated for the squirrel's terminal velocity.

So, the velocity of the 58.0-kg person hitting the ground, assuming no drag contribution, would be the same as the squirrel's terminal velocity.