What force must a person exert to jump 75 cm high

  • Thread starter Thread starter Woofuls
  • Start date Start date
  • Tags Tags
    Force Jump
Click For Summary
SUMMARY

To determine the force a 75kg person must exert to jump 75 cm high, the maximum potential energy (PE) is calculated using the formula PE = mgh, resulting in 225J when the person crouches 30 cm. The required force is derived from the equation Work = force * mass, leading to an initial force of 500N. However, considering the acceleration needed to achieve the jump height, the total force required is 1250N. This analysis incorporates both kinematic equations and energy methods to arrive at the solution.

PREREQUISITES
  • Understanding of potential energy (PE) calculations
  • Familiarity with Newton's second law (F = m * a)
  • Knowledge of kinematic equations for vertical motion
  • Basic grasp of energy conservation principles
NEXT STEPS
  • Study kinematic equations for vertical motion in-depth
  • Learn about energy conservation in mechanical systems
  • Explore advanced applications of Newton's laws in sports science
  • Investigate the biomechanics of jumping techniques
USEFUL FOR

Physicists, sports scientists, coaches, and anyone interested in the mechanics of jumping and athletic performance optimization.

Woofuls
Messages
4
Reaction score
0
<i>An exceptional jumper could reach a vertical height of 0.75m. What force must a 75kg person exert to achieve this height? (assume the person crouches 0.30m to jump).</i>

Okay, I know that what I'm going to have to start with:
y = viy * t + 0.5 * g * t^2

to find max height at some velocity...
(.75m-.30m) =d (10m/s^2)= g
.45m = (viy) * t = .5 * g * t^2

Right, so I have no time, but there is:
t = (vi)/g

Hrm, I still don't have (vi) or t...

My next thought is to find maximum PE...

PE = mgh

PE = 75kg * 10m/s^2 * .3m = 225J

So, I know max energy...

Work = force * mass
225J = f * 75kg, f = 500N

So, I know he can push off with a max of 500N

500N = 75kg * a
F = m * a
a = 6.66m/s^2

so I can use
6.66*t= .45m
t=.0676secs

y = viy * t + 0.5 * g * t^2
y = (6.66) * .0676 = .5 * 10 * .0676^2
about .47m, so I'm very close!

F = m * a
F = 75kg * 6.66 =500N + 750N = 1250N required

I know this is wrong, I would appreciate some help:cry:

Thanks in advanced!
 
Physics news on Phys.org
Woofuls said:
An exceptional jumper could reach a vertical height of 0.75m. What force must a 75kg person exert to achieve this height? (assume the person crouches 0.30m to jump).
Some hints on how to understand this problem:

I assume that the "vertical height" of 0.75m means that the person's center of mass rises that much above its usual position (when the person is standing). In other words, after the person leaves the ground, he rises an additional 0.75m.

I assume that the "crouches 0.30m" means that the ground only applies a force on the person while his center of mass moves a distance of 0.30m. (That's the force you are asked to find. But don't forget gravity.)

There are several ways to solve this problem. You can use kinematics: Find the person's speed the second he leaves the ground, then find his acceleration during his jump, etc. Or you can use energy methods.

Give it another shot.
 

Similar threads

Replies
1
Views
3K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 41 ·
2
Replies
41
Views
5K
Replies
8
Views
3K
  • · Replies 15 ·
Replies
15
Views
6K
Replies
1
Views
4K
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
23K
  • · Replies 4 ·
Replies
4
Views
2K