Question

a 50 kg pole vaulter drops vertically from a height of 4 meters coming to rest just .4s after his feet touch the ground, what is his momentum just before his feet touch the ground? what net force acts on him during his acceleration?

Answer 1

1) His momentum before touching the ground is 442.5 kg m/s

2) The net force on the pole vaulter is 490 N

Explanation:

1)

The motion of the pole vaulter is a motion of free fall, so it is subjected only to the force of gravity. Therefore, he falls to the ground at constant acceleration ($g=9.8 m/s^2$, acceleration of gravity), so we can use the following suvat equation to find his final velocity:

$v^2-u^2=2as$

where

v is the final velocity

u = 0 is the initial velocity

$a=g$ is the acceleration

s = 4 m is the distance covered during the fall

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(4)}=8.85 m/s$

And therefore, his final momentum before touching the ground is

$p=mv$

where m = 50 kg is his mass. Substituting,

$p=(50)(8.85)=442.5 kg m/s$

2)

As we said previously, the only force acting on the pole vaulter during his fall is the force of gravity, which is

$F=mg$

where

m = 50 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting, we find the force of gravity, and therefore the net force acting on the pole vaulter:

$F=(50)(9.8)=490 N$

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