Please help me on this physics question! (Involves momentum)?

Question

A 55 kg pole-vaulter falls from rest from a height of 5.0 m onto a foam-rubber pad. The pole-vaulter comes to rest 0.30 s after landing on the pad.
a) Calculate that athlete's velocity just before reaching the pad.

b) Calculate the constant force exerted on the pole-vaulter due to the collision.

Answer 1

To find the velocity just before the athelete hits the pad:

mgh = 1/2 m v² so v = sqrt ( 2gh )

Now that you have velocity, compute momentum mv.

Now impulse = Ft = change in momentum

F ( .30 s ) = mv

Solve for force.

Answer 2

It does not have angular momentum approximately its possess axis, however any relocating item, despite the fact that it's relocating in a flawlessly instantly line and isn't rotating, can nonetheless have angular momentum. It simply depends upon your body of reference. Imagine you're jogging in the direction of a merry-cross-circular (which is not relocating). If you run instantly on the core of it, you are going to simply crash into it and not anything extra will occur. BUT should you run on the outer rim and leap on, the MGR will begin to rotate. You have given it a few angular momentum. Angular momentum is a conserved volume - it cannot come from nowhere, so that you ought to have had a few earlier.... The change within the 2 instances is that, within the moment illustration, you had a distance among the axle and your speed. The perpendicular distance among speed and the reference factor is the primary element. The angular momentum (approximately a reference factor) of a unmarried item with mass 'm' and speed 'v' at a perpendicular distance 'R' from the reference factor is L=mvR So to make any declaration approximately whether or not an item has angular momentum or now not, you have got to state the reference factor.

Answer 3

a)m*g*h=0.5*m*v^2 solve v..
b)m*g*h=0.5*m*v^2 +F*(v*0.3) solve F..