Falling on a handrail

Question:

A 68kg person falls against a handrail. With what force, in N does the person hit the handrail? Assume that the person falls 0.6m, trips and falls against the handrail.

Answer:

Take the acceleration due to the Earth's gravity to be 9.8 m/s².

Does the person fall straight down or at and angle?
How long do they take to stop?

Answers to both these questions are essential if you are going to work out the force on the handrail.

I'll give you some examples, you can always alter the numbers if they are not what you want as long as you follow the principle of the argument.

1. Falling straight down and stopping in 0.05 s
Velocity after falling 0.6 m use v² = u² + 2as where s is the height fallen, v the final velocity and u the initial velocity (=0 if we assume that the person falls from rest).

Putting the numbers in gives v = 3.43 m/s
Momentum that they lose when hitting the rail = mv = 68 x 3.43 = 233 Ns

Force on stopping = [momentum change]/[time of stopping] = 233/0.05 = 4664 N

2. Falling and hitting the handrail at an angle of 60 degrees to the horizontal and stopping in 0.1 s. To do this they must have been given a sideways push – lets assume it gave them a horizontal velocity of 0.4 m/s. This horizontal velocity will be constant until they hit the rail because there is no horizontal acceleration for objects fall at an angle. I will also assume that 60 degrees is the angle at which the HIT the handrail.
The vertical velocity with which they hit will be the same as before if the fall vertically 0.6 m = 3.34 m/s.
But this time they also have a horizontal velocity of 0.4 m/s and so the total velocity (Pythagoras) = 3.45 m/s

So using the same equations as before the force = 68x3.45/0.1 = 2346 N.

3. What if they just topple sideways? This would result in them not really falling but just leaning on the rail. In this case it matters how much their centre of mass changes height as they lean.