How does inertia work with gravity and kinetic energy?

Question

If a person is on a moving train going 150 mph, then jumps out and falls about 60 ft. into water, how fast is that person moving when they break surface tension? This question is from the final action sequence in XXX 2: State of the Union where Ice Cube jumps out of the bullet train, and the above mentioned questioned was raised.

Answer 1

There are several forces at work here.

Horizontal: drag due to air friction slowing the horizontal velocity down
Vertical: weight, speeding the vertical velocity up

The total velocity when breaking surface tension is the vector sum of the horizontal (vh) and vertical (vv) velocities = V = sqrt(vh^2 + vv^2). The vertical velocity = vv = gt = g sqrt(2S/g) = sqrt(2Sg); where S = 60 ft and g = 32.2 ft/sec^2.

We can't find vh with what you've given us because the initial horizontal velocity vh0 = 150 mph will be slowed down by the drag friction. And drag friction forces depend on air density, coefficient of drag, and the cross sectional area of the person...none of which has been given.

However, one rule of thumb in such matters is that the terminal velocity of a person falling through the atmosphere is about 120 mph. So we can suppose that vh1, the end horizontal velocity, will be somewhere between 120 and 150 mph. Let's assume the soft landing and vh1 = 120 mph = 176 ft/sec as the terminal horizontal velocity.

Thus, the break surface tension velocity V = sqrt(vh^2 + vv^2) = sqrt(176^2 + (2Sg)^2) = sqrt(31000 + 3800) = 186 ft/sec or about 127 mph. Ouch. At those speeds, water tends to act like concrete...in other words, splat...end of XXX.

Answer 2

Just roughly you can consider the forward motion (from being on the train) seperately from the downwards motion from gravity (this is assuming the train was on a horizontal path, not heading slightly down or up). For the forwards motion the person will keep this but it will be reduced a bit because of the body's drag while travelling through air. 150mph=67m/s so we can guess the person has maybe 50 or 60 m/s forwards velocity left by the time they hit the water.

For the downwards motion they will gain a bunch of velocity which you can figure out using

vf²=vi²+2ad

vf=final velocity, vi=initial velocity, a=acceleration due to gravity, d=distance

Since we are considering only downwards motion at this point we set the vi as 0, a as 9.8m/s² down, d as 18 meters (that's about 60 feet)

vf²=0+2(9.8)(18)
vf=18.8 m/s [down]

The person won't be going quite that fast again because of drag from going through air.

If you add the forwards and downwards velocity vectors you get that the person is moving at about 50 to 60 m/s mostly forwards but heading about 15 degrees down from horizontal when they hit the water. So you would lose velocity from drag and gain velocity from acceleration due to gravity and still end up hitting the water way too hard, not a whole lot less then 150 mph. (This is rough because I fudged and guessed a bit with the drag involved but it shouldn't be too far off the mark).

Answer 3

In this problem we can likely neglect any air friction or other influencing factors. Therefore the person's horizontal velocity will stay constant during the entire problem and will be equal to the horizontal velocity of the train (150 mph, 67 m/s).
As the object falls and accelerates due to gravity, it gains vertical velocity. Since we can assume that g = 9.81 m/s^2 and we know the fall distance to be 60 ft (18.3 m), we can then calculate the change in the vertical velocity.
Using the concept of conservation of energy, the person's initial gravitational potential energy is converted into kinetic energy,
PE = KE = mgh = 1/2 mv^2
v = sqrt (2gh) = 18.95 m/s (downward).
This is the final velocity in the y direction, we must now add it to the x component of th velocity,
x^2 + y^2 = v^2
where x and y are the horizontal and vertical components of the person's velocity respectively, and v is the total velocity.
Here, as the person hits the water below, their total velocity is,
67^2 + 18.95^2 = v^2
v = 69.6 m/s

For conversion to and from SI units,
http://www.digitaldutch.com/unitconverter/