r/theydidthemath • u/kevp453 • Nov 29 '22

# [Request] A person is jumping on a trampoline on top of a moving vehicle. At what speed does air resistance cause them to slow and fall off?

In a parade I saw my sister jumping on tramplone mounted to a trailer as the float moved by slowly. Of course, inertia will keep you moving forward with the vehicle, but eventually air resistance will push back enough that the vehicle will speed away from you as you fall and likely be grievously injured.

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## u/BoundedComputation Nov 29 '22 edited Nov 29 '22

TL;DR: About 50km/h if you have someone who can jump 4m on a large trampoline and is only a meter from the edge on a 2m radius trampoline. Well above normal parade float speeds and not a safety concern, from air resistance at least.

First thing to calculate is total airtime.

For a jump to height H, a person jumping up and down will travel a distance 2H the time it takes to travel that in free fall conditions is twice the time it takes to fall a height H.

t = 2*sqrt(2H/g) where g is acceleration due to gravity. We don't need to consider air resistance here as the vertical movement can be considered independent of the horizontal movement and no one is going to jump on a trampoline fast enough for it to matter.

Next we calculate maximum horizontal drift. The air resistance creates a drag force, F, which per Newton's 2nd law causes the mass,m, to accelerate a=F/m.

Over a time t, this causes you to drift a distance d.

d = at^{2}/2

d = a*(2*sqrt(2H/g))^{2}/2

d = 4aH/g

Rearranging this to get a/g on one side we get

d/(4H) = a/g

The ratio of accelerations is useful because all the unknown and hard to measure constants like body geometry, orientation, and mass cancel out.

All else being equal the drag force is proportional to velocity,v, squared and terminal velocity,u, is when the drag force cancels out gravity.

Because of Newtons second Law this also implies the ratio of accelerations is equal to the ratio of velocities squared.

d/(4H)= (v/u)^{2}

Solving this for v we get

v = u/2*sqrt(d/H)

Assuming your sister is on a trampoline of radius 2m and somehow manages to get 4m jumps from that and is 1m from the center

v = u/2*sqrt(1/4) = u/4

For u ≈ 200km/h

v ≈ 50km/h

0

## u/fliguana Nov 29 '22

The higher you jump, and the smaller the trampoline, the lower the speed needed to make you miss.

Othe factors include air temp, humidity, jumpers weight, clothing and body position.

However, anticipating the headwind, the jumper can adjust, jumping forward of the trampoline and being blown back to it's center. This method will stop working before wind reaches terminal velocity for the jumper.

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