Solving Paratrooper's Fall: Estimating W, F, & Air Resistance

AI Thread Summary
The discussion revolves around calculating the work done by snow on a paratrooper who fell 370 m and landed in a snowbank, as well as the average force exerted by the snow and the work done by air resistance. The initial calculations for work and force were incorrect, prompting a reevaluation of the concepts of work, energy, and Newton's laws. It was emphasized that the net work done on the paratrooper equals the change in kinetic energy, which includes contributions from both gravity and the snow. The importance of understanding conservation of energy and the correct application of work formulas was highlighted. Accurate calculations require a clear grasp of these principles to arrive at the correct answers.
cyspope
Messages
27
Reaction score
0

Homework Statement



71. A paratrooper fell 370 m after jumping from an aircraft without his parachute opening. He landed in a snowbank, creating a crater 1.1 m deep. Assuming the paratrooper's mass was 80 kg and his terminal velocity was 30 m/s, estimate : a) the work done by the snow in bringing him to rest; b) the average force exerted on him by the snow to stop him; c) the work done on him by air resistance as he fell.

A. a) -36000 J , b) -3300 N , c) -250000 J

Homework Equations


W = mgh = Fdsin\theta= \frac{1}{2}mv^{2}


The Attempt at a Solution


a) I got a) by using .5mv^{2}. However, I don't know how to get b) and c)
 
Physics news on Phys.org
what's making the paratrooper fall? Force by gravity. So to stop him what force must be applied? Think of Newtons third law.

What is work? work is force times distance. The distance is the depth of the crater, what is the work?
 
Where are these answers coming from? None are correct...
 
Newton's third law is about action and reaction, isn't it?

this is what i did for b)
W = Fdsin\theta = 80 * 9.8 * 1.1 = 862.4 Joule

However, this is wrong because my answer is different from the given answers.
 
cyspope said:
Newton's third law is about action and reaction, isn't it?

this is what i did for b)
W = Fdsin\theta = 80 * 9.8 * 1.1 = 862.4 Joule

However, this is wrong because my answer is different from the given answers.

your answer and the given answers are incorrect. The net work done on the 'trooper in bringing him to rest is his change in KE. The net work includes the work done by the snow and the work done by gravity. Once you calculate the work fone by the snow, then the force of the snow on the trooper can be calculated using the definition of work. Are you familiar with conservation of energy equations? Otherwise, you'll have to use the kinematic equations and Newtons laws.
 
What is work?

Work=\int F ds

From this you can find that

Work=-\Delta U and Work=\Delta KE

Read your book about work and energy.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top