__The Falling Pedestrian (more correctly The Toppling Pedestrian)__

by Eddie Baart

This problem was set by Duane Meyers. He wrote as follows.

"I would like your thoughts on a possible homicide involving a pedestrian who fell from a ledge at
a parking ramp, down to a sidewalk below.

**Facts**

The top of the ledge is a vertical distance of 44.0 feet from the ground. The ped was found
approximately 12 feet horizontally from the edge of the ledge. His CM is believed to be
approximately 35 inches from the bottom of his feet.

__Questions__

What would the ped's horizontal velocity be if you assumed that the ped's body were rigid from the feet to the CM and he tipped out away from the ledge while standing on it ?

What would be the probable horizontal distance traveled from the ledge to the point of impact ?"

__The Solution__

Suppose the pedestrian is of height *l *(*l/*2 is distance from
center of gravity to his feet). We describe the motion of the
center of gravity G in terms of the coordinates 2 and *r* as
shown in the diagram. The *x* coordinate of G is *l/*2 cos q
so its *x* component of velocity (by differentiation) is

and its *x* component of
acceleration is

...............(1)

The two terms in the *x* acceleration can be recognized as the horizontal components of the
tangential acceleration and the radial (centripetal) acceleration .

There is a radial acceleration as in uniform circular motion because, although is not changing in magnitude, its direction is changing.

The feet of the pedestrian will lose contact with the ledge when the horizontal component of the
force exerted on them by the ledge (**H** in the diagram) becomes zero i.e. when the *x* component of
the acceleration shown above becomes zero.

In order to find the angle at which this occurs we need to find expressions for and . To do this we use Newton's Second Law (for rotation) and Conservation of Energy.

**Newton's Second Law**

For rotational motion this law states that the moment of inertia times the angular acceleration
equals the torque ( by analogy with **ma = F** for linear motion). The moment of inertia of the
pedestrian for an axis through his feet is 1/3 *l*^{2 }. So

and hence ....................................(2)

**Conservation of Energy**

The rotational kinetic energy when the pedestrian is at an angle must equal the gravitational
potential energy it has lost as his center of mass (CM) descended from its previous height of *l*/2
above his feet to the new height *l*/2 cos q. Thus

and so ........................(3)

**The angle of launch**

We now use the condition for the horizontal force to be zero i.e. that the *x*** **acceleration of
equation (1) is zero.** **

If we substitute equations (2) and (3) into equation (1) and put , we get and
hence It is noteworthy that this result is independent of *l* and so it applies to the
pedestrian as well as to a pencil toppling off a table.

**The components of launch velocity**

We can get the horizontal component of his velocity at the instant his feet leave the ledge by substituting into the equation (3) to find and then using the first equation on page 1.

This gives ...........................(4)

and we can get a similar expression for the vertical speed when

.........(5)

**Duane's Problem**

We can apply equations (4) and (5) to Duane's problem with *l* = 2 x 0.889 m (2 x 35 inches).

The height of the CM at launch is *l*/2 cos q above his feet = 0.593 m and vertical velocity is 1.557
m/s.

So it falls 0.593m + 13.41 m (44 ft) = 14.00 m and this takes 1.54 seconds.

Since the horizontal launch velocity is 1.391 m/s he travels a horizontal distance of 2.14 metres
(7.03 ft).

If we compare this with Duane's distance of 12 feet, then it is obvious that this was not a simple
topple with the assumptions I have used. Whether foul play occurred, I leave you to judge.

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