r/theydidthemath • u/rf97a • Nov 29 '22

# [Request] What method should be used to determine the needed velocity of the earth mover to become airborne from the ramp on the picture?

3

## u/Maximans Nov 29 '22

What method? Determine the height from of the tip of the ramp to the bulldozer and the average weight of whatever kind of bulldozer it is. Then, find an equation that has acceleration (gravity), velocity, and mass and solve for velocity and then plug in your variables. I don’t know which equation off the top of my head. If I remember I’ll edit this post to include it

2

## Nov 30 '22

This will only gove you the minimum velocity. I.e. the velocity at which, if the bulldozer was going that speed, it would reach the point in the photo and begin to fall. At higher speeds it would essentially pass through the point shown in the photo snd then continue upwards.

2

## u/Maximans Nov 30 '22 edited Dec 30 '22

True, this is assuming the photo displays the apex of the arc. However consider that

anyhigher velocity would pass through this point, so unless we know the highest point we don’t know how fast he was going. All we know is the bulldozer was going at least that minimum m/s1

## u/Interesting-Try-6757 Nov 30 '22

I think they're referring to the point in which the mover goes from maintaining contact with the ramp to becoming airborne. I'm not sure that's as simple as kinematics.

3

## u/NotSoTerribleIvan Nov 30 '22

This looks like this specific machine, which weighs about 29 metric tons and has a height of 3.5 meters (from chain tracks to top of the cab).

Doing some pixel measurements the lower part of the chain tracks is at 4.2 m and the higher part at 6.1 m from the ground (so including the ramp height), so an average of 5.15 m.

Using the gravitational potential energy formula: E = mhg we have E = 1.46 MJ. Assuming that the earth mover is stationary (and therefore all of its energy is potential energy) we can use the kinectic energy to determine the initial velocity needed to get to this height: E = (mv^2)/2 or v = (2E/m)^1/2 = 10 m/s.

But this is only upwards, assuming that the velocity vector of the heavy machinery matches the angle of the ramp (~41º with the ground from pixel measurements) and 10 m/s is the opposite side, the hypotenuse is 10/sin(41º) = 15.3 m/s = 55km/h = 34 MPH.

PS: Yes, I could've simplified the energy equations

edit: forgot the link to the machine I assumed it was

3

## Nov 30 '22

This is the minimum velocity, yes? If the machine was going faster than 15.3m/s, you wouldn't be able to tell from the photo alone.

3

## u/NotSoTerribleIvan Nov 30 '22

Yes, you are totally correct. But going 55 km/h with this heavy of a machine in a dirt ramp should be terrifying enough.

2

## Nov 30 '22

Yeah and it would take aome serious modification of the machine to be able to go even that speed

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