## To the editors:

Cecil Adams's answer about which bullet hits the ground faster [Straight Dope, December 16] is right, but for the wrong reasons. There is a much stronger reason than curvature of the earth why the dropped bullet hits the ground faster than the fired one, namely air resistance.

Air resistance is a force which tends to slow down an object, and its strength is roughly proportional to the square of the speed of the object. A dropped object takes quite a distance to get up speed (the formula is speed equals the square root of 2 x g x distance, where g is gravity and is 32 ft/sec(2), so after dropping 100 ft, an object would be going the square root of 6400 or 80 feet per second, not counting air resistance). On the other hand, the fired bullet starts out really fast--probably about 1000 ft/second.

So what is the effect of air resistance? On the fired bullet, it slows it down rapidly, say by 50% within a few seconds. That reduction of speed affects both the horizontal and vertical components of the trajectory, so that especially the vertical component of the motion is only 50% what it would have been--i.e., the bullet is only dropping at half the rate without air resistance.

On the other hand, the bullet which is dropped has minimal slow down due to air resistance because it's not going very fast. So it almost exactly obeys the formula (dist.) = 1/2 gt(2) where t is elapsed time. The fired bullet, however, is obeying a formula more like (vertical dist.) = 1/2g(1) squared where g(1) is the effective vertical acceleration and is considerably less than g (gravity) because of air resistance.

I'm not sure you want to try to explain this to your readers, but good luck making a coherent explanation if you do. By the way, I'm not a physicist, but I did take quite a lot of physics in college.

Alfred Magnus

New York, New York

## Cecil Adams replies:

What college, Acme School of Beauty? Your explanation is a gussied-up version of the common sense (i.e., wrong) answer. The arc described by the bullet shot from the gun is a function of two forces, the exploding gunpowder and gravity. Wind resistance affects the two independently. Mathematically we know this because when we calculate the distance the bullet travels before it hits the ground, we are obliged to plug wind resistance into two different places in the equation. The downward velocity, which may be crudely expressed as "gravity minus wind resistance," is mathematically identical in the case of both the fired bullet and the dropped bullet. The time it takes the fired bullet to fall to the ground vertically is of course the limiting factor in how far it travels horizontally. Ergo, the fired bullet and a dropped bullet hit the ground at the same time--ignoring, of course, the curvature of the earth. Don't believe me? Too bad. This well-known problem is straight out of Physics 101.

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