How far can a 55kg person fall and break and leg?

In summary: Power is measured in watts (w). In summary, Cohort explains that a person can jump up to a certain height and land upright, but if they weigh more than 55 kg, they may break their legs. He also provides an equation to determine the energy a human leg can absorb. He is stuck on one last question and figures out the power it would take to move an elevator at a constant velocity. His units are wrong, but the rest of the calculation sounds correct.
  • #1
cohort
6
0
Hi, having a problem with this question. I thought it was as simple as the conservation of energy, but I guess there is something I'm missing. The program we're using isn't accepting my answer, even when I use two significant digits.

Thanks for the help,

Cohort

Homework Statement


From approximately what maximum height could a 55 kg person jump and land rigidly upright on both feet without breaking his legs?

The energy a human leg bone can absorb is 200 J.
The person is 55 kg.

Homework Equations



mgy = 1/2mv^2

The Attempt at a Solution



(55 kg) (9.8 m/s^2) y = 200 J
y = 200 J/ 539
y = .37 m
 
Last edited:
Physics news on Phys.org
  • #2
cohort: Generally always maintain four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits, unless the first significant digit of the final answer is 1, in which case round the final answer to four significant digits. Have you tried this yet? Note also that numbers less than 1 should always begin with a zero before the decimal point. Omitting the zero violates ISO 31-0. How many tries do you have remaining? Your answer looks correct to me, so far, unless I am missing something.
 
  • #3
I've tried so many times I might not get credit for it, but I don't want to choose "show answer" in case I can still get credit for it.

55 kg * 9.8 m/s^2 = exactly 539 N. 200 J / 539 N = .3710575139 They want two significant digits so I put in 3.7*10^-1, but it still said Try Again. This is frustrating.

Should I just hit Show Answer, or is there some other way of entering the height I haven't tried yet?

Thanks for the help.

Edit: The zero thing didn't work either.

By the way, I'm using mastering physics.
 
  • #4
cohort: No, don't hit "show answer" yet. Let us think it over awhile. When is the deadline? Wait until just before the deadline. Have you tried entering 0.37 and 0.371 yet? Regardless, don't hit "show answer" yet if you don't have a deadline yet.
 
  • #5
Midnight. Its one of the last two problems I have left.

I tried 37 * 10^-2 also, and it didn't work.
 
  • #6
cohort: Have you tried entering 0.37 and 0.371 yet, without scientific notation? Secondly, are units also required by the program when inputting an answer, such as 0.37 m or 0.371 m? How many hours from now is midnight?
 
  • #7
I tried both of those. Neither worked. Units are automatically included in the answer.

It is 10:29 where I am.

I'm thinking I'll just e-mail my teacher and tell him the program is wrong. It has to be!
 
  • #8
Wait an hour before doing anything. Let us think it over awhile.
 
  • #9
Hint: How many legs does a typical human have?
 
  • #10
2. Oh, wait so its twice as high!

Haha. Thank you so much. It was the little details that got me on that one. Just one more problem to go and I'm done!
 
  • #11
Did it give you credit? Or did it say you had too many tries?
 
  • #12
I got credit. A little under half, but its better than nothing. Thanks again!

If you could, I'm stuck on this last question too. Its finding the power it takes to move a 2300 N elevator at a constant velocity of 7.0 m/s.

Work = force x distance. I figure 2300 N over 7 m = 16100 J.

Power = work over time, so 16100/ 1 second = 16100 kw.

I sounds right to me, but am I missing something?

Edit: BrB, drive home.
 
Last edited:
  • #13
If this helps, the units on your final answer are wrong.
 

Related to How far can a 55kg person fall and break and leg?

1. How does the weight of a person affect their risk of breaking a leg when falling?

The weight of a person does play a role in their likelihood of breaking a leg when falling. Generally, the heavier the person, the greater the impact and force when they hit the ground. However, other factors such as the surface they are falling on and their body's positioning can also influence the risk of injury.

2. Can a 55kg person break their leg from any height?

Yes, a 55kg person can potentially break their leg from any height. The force of impact and the positioning of the body upon landing are important factors in determining the likelihood of injury. Even a fall from a short height can result in a broken leg if the person lands in an unfavorable position.

3. What is the typical height at which a 55kg person would break their leg from a fall?

There is no specific height at which a 55kg person would break their leg from a fall. The risk of injury depends on various factors such as the surface they are falling on, their body's positioning, and the force of impact. However, falls from higher heights are generally more likely to result in injury.

4. Can a 55kg person minimize their risk of breaking a leg when falling?

Yes, there are some steps a person can take to minimize their risk of breaking a leg when falling. These include maintaining good balance and body control, trying to land on a soft surface if possible, and protecting the legs by positioning them in front of the body upon impact.

5. What should a 55kg person do if they break their leg from a fall?

If a 55kg person breaks their leg from a fall, they should seek medical attention immediately. Depending on the severity of the injury, they may need to immobilize the leg and use crutches or a wheelchair while it heals. Physical therapy may also be recommended to aid in the recovery process.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
4K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
5K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
3K
Replies
1
Views
924
  • Introductory Physics Homework Help
Replies
4
Views
2K
Back
Top