Standing broad jump on the moon

  • Thread starter Thread starter nemesis08
  • Start date Start date
  • Tags Tags
    Jump Moon
AI Thread Summary
The discussion revolves around understanding the physics of a standing broad jump on the moon, specifically focusing on the force generated during the jump. The average force is given as F = 2W, which raises confusion about the units, as W typically refers to power rather than force. Participants emphasize the necessity of knowing the jumper's mass to accurately compute the jump's range, maximum height, and duration. They highlight the importance of considering the force applied over a distance and its relation to mass and velocity. The conversation underscores the challenges faced due to unclear explanations from the professor.
nemesis08
Messages
3
Reaction score
0
i am having trouble with this exercise, i don't understand much and the professor doesn't explain too well so I am kinda lost.



Consider a person on the moon who launches herself into a standing broad jump at 45 degrees. The average force generated during launching is, F = 2W and the distance over which this force acts is 60 cm. kindly compute:

a. The range of the jump
b. The maximum height of the jump
c. The duration of the jump
 
Physics news on Phys.org
I'm not sure I understand what F=2W means. W could be watts but that's not a force that's a power unit. If you meant 2N as in Newtons then...

You still don't have enough information to solve the problem. You need the person's mass...consider if you push a 10 metric ton bus with a measly 2N over 60cm it is barely going to go anywhere but if you use a slingshot to do this with a small rock then you'll get a really long and high trajectory. Mass matters.

But assuming you have the mass then note that they give force and distance over which it acts. What quantity is force times distance? How does this relate to mass and velocity?

Also consider potential energy on the moon. Remember its surface acceleration is quite close to 1/6 Earth's 9.8 m/s^2.
 
thanks tho, i don't know that's what the exersice say. That profesor is crazy and i really don't understand anything.


Lets see what i can do
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top