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?
To find the velocity of a 58.0-kg person hitting the ground with no drag contribution, you can use the principle of conservation of energy.
The potential energy of the person at the top of the fall is given by:
PE = m * g * h
Where m is the mass of the person, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height from which the person falls.
Assuming no drag contribution, all the potential energy is converted to kinetic energy at the bottom of the fall. The kinetic energy is given by:
KE = (1/2) * m * v^2
Where v is the velocity of the person at the bottom of the fall.
Equating the potential energy to the kinetic energy, we have:
PE = KE
m * g * h = (1/2) * m * v^2
Canceling out the mass (m) on both sides of the equation, we get:
g * h = (1/2) * v^2
Solving for v, we have:
v = √(2 * g * h)
Plugging in the values, with g = 9.8 m/s^2 and h being the height from which the person falls, you can calculate the velocity of the person hitting the ground.
To calculate the velocity of a 58.0 kg person hitting the ground assuming no drag contribution, we can use the principle of conservation of energy.
First, let's assume that the person falls from the same height as the squirrel, which is 5.2 m.
The potential energy (PE) of the person at the initial height is given by:
PE = m * g * h
where
m = mass of the person = 58.0 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height = 5.2 m
PE = 58.0 kg * 9.8 m/s^2 * 5.2 m
PE = 2989.6 joules
This potential energy is converted to kinetic energy (KE) as the person falls:
KE = 0.5 * m * v^2
where
v = velocity
We can equate the potential energy to the kinetic energy:
PE = KE
2989.6 joules = 0.5 * 58.0 kg * v^2
Dividing both sides by 0.5 * 58.0 kg:
v^2 = 2989.6 joules / (0.5 * 58.0 kg)
v^2 = 102.7 m^2/s^2
Taking the square root of both sides:
v ≈ √(102.7) m/s
v ≈ 10.14 m/s
Therefore, the velocity of a 58.0 kg person hitting the ground assuming no drag contribution would be approximately 10.14 m/s.
v^2 = 2gx
solve for v