Calculate the impact force when falling from a height

[Emilyyyyyyyyy]

Homework Statement

I have this scenerio where a man A jumps from a 400 m building. He is caught 5 meters below by man B (so at 395 m). man A hits with man B given his mass is 79.4 kg and the horizontal distance travelled is 3.86 meters, all in 1.5 seconds. How do I calculate the force that man A hits man B with?

Relevant Equations

vx = dx/t
W = FnetΔd

To find vx
vx = dx/t = 3.86 m/1.5 s= 2.573 m/s

To find Ek
Ek = ½mvx²= ½(79.4)(2.573)²= 262.8 J

W = FnetΔd
Fnet = 262.8 J/ 3.86 m = 68 N

He hits him with a force of 68N

 

[haruspex]

You need to explain the set-up more clearly.

Whose mass is 79.4kg? Do we know the mass of the other?

What is this horizontal distance? Is that from jumping to being caught, or is there further movement, maybe with both men together? What, exactly, takes 1.5s?

You won't be able to find an actual force. You could perhaps find an average force if you knew how long the impact lasted, but it is surely a lot less than 1.5s.
Do you mean you want to find the impulse?

 

[CWatters]

+1

The force depends on how fast man A decelerates after hitting man B. This cannot be calculated using the information provided.

If man B is 2m tall then it might be reasonable to assume the stopping distance of man A is less than 2m. This could be used to estimate the deceleration and the minimum force of impact.

However the maximum force could be much higher.

 

[osilmag]

The falling man is going to be moving much faster in the y direction when he impacts the man on the ground than 2.573 m/s.

$vf = v0 + gt$

 

[neilparker62]

The man falls under the influence of gravity. So you can calculate the time Δt it takes to fall 5m and deduct it from 1.56 s to determine impact time. You can also use Δt to determine the horizontal and vertical components of the man's velocity and hence a resultant momentum. Assuming the 'catch' takes up the remaining time, the force (or average force) may be determined as Δp/(1.56 - Δt). Another assumption would be that the given horizontal and vertical distances are pre-collision.

 

[jbriggs444]

The man falls under the influence of gravity. So you can calculate the time Δt it takes to fall 5m and deduct it from 1.56 s to determine impact time. You can also use Δt to determine the horizontal and vertical components of the man's velocity and hence a resultant momentum. Assuming the 'catch' takes up the remaining time, the force (or average force) may be determined as Δp/(1.56 - Δt). Another assumption would be that the given horizontal and vertical distances are pre-collision.

We are not told that the jump is horizontal. The time for the fall cannot be determined in this manner.

In addition, even if we grant the assumption of a horizontal launch, a fall of 5 meters would take about 0.7 seconds leaving 0.8 seconds for the collision 1.0 seconds leaving 0.5 seconds for the collision. It might be plausible for a high acceleration catch (over a distance of no more than about 2 meters) to match with a presumed to be instantaneous jump and a 9.8 m/sec² fall (over a distance of 5 meters).

Taking the 1.5 second figure at face value would allow one to compute the required vertical component of the launch velocity to arrive 5 meters below at the appointed time. It would also allow one to compute the required horizontal component of the launch velocity to arrive 3.86 meters out at the appointed time. With launch velocity in hand, one could proceed to determining impact velocity and impact momentum.

That leads straight to the problem that @haruspex pointed out in #2 -- without a time taken for the collision there is no way to convert impact momentum to the average force required to dissipate it.

Edit: Thank you, @neilparker62 for the sanity check on the fall time.

 

[neilparker62]

Once again one is guessing the situation. I am imagining a person running along the top of a building (presumably on fire or something) and jumping towards rescuers in a nearby building. Otherwise how else can there be horizontal motion at all ? I am also imagining the person to be jumping as 'flat' as possible - hence horizontal launch and zero initial vertical velocity.

5 = 1/2 x 9.8 x t^2 gets me to t=1.01 s leaving 0.49 s for impact time (?). In my previous post, I did misread 1.5 s as 1.56 s for some reason.

Perhaps the OP could clarify for us the exact situation as requested by Haruspex in post #2 ?

 

[jbriggs444]

5 = 1/2 x 9.8 x t^2 gets me to t=1.01 s leaving 0.49 s for impact time

Right you are. I was taking an inappropriate square root of 0.5.

 

[neilparker62]

Something like this perhaps: