How Much Force Stops a Pole Vaulter?

[sensesfail]

Homework Statement

a 58kg pole vaulter falls from rest from a height of 5.3m onto a foam rubber pad. the pole vaulter comes to rest 0.38 seconds after landing on the pad

Calculate the constant force exerted on the pole vaulter due to the collision. Answer in N

Homework Equations

d = Vi(t) + (1/2)(a)(t²)
Vf = Vi + (a)(t)

The Attempt at a Solution

d = Vi(t) + (1/2)(a)(t²)
5.3 = 0 + (1/2)(a)(.38²)
73.4 m/s² = a

Vf = Vi + (a)(t)
Vf = 0 + (73.4)(.38)
Vf = 27.89 m/s

 

[rl.bhat]

d = Vi(t) + (1/2)(a)(t²)
5.3 = 0 + (1/2)(a)(.38²)
73.4 m/s² = a This is wrong. You have to find the velocity with which he lands on the rubber pad. That is his initial vlocity. And his final velocity is zero..Time is given. Find the retardation and hence the force.

 

[ogb p]

To calculate the force exerted on the pole vaulter, we can use the equation F = ma, where m is the mass of the pole vaulter (58kg) and a is the acceleration (73.4 m/s²). Therefore, the force exerted on the pole vaulter due to the collision is:

F = (58)(73.4)
F = 4257.2 N

This means that the pole vaulter experienced a force of 4257.2 Newtons during the collision with the foam rubber pad. This force was enough to bring the pole vaulter to a complete stop in just 0.38 seconds. It is important to note that the actual force experienced by the pole vaulter may be slightly different due to factors such as air resistance and the compressibility of the foam rubber pad.