# »  Solutions to puzzles in my National Review Online Diary

## May 2010

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In my May diary I posed the following brain-teaser.

Here are specs for the L115A3 long-range sniper rifle, the one Cpl. Craig Harrison used for that astounding feat of marksmanship we read about earlier in the month.

Question:  Using basic trig and the usual equations of motion, assuming sniper and target both at ground level, and ignoring air resistance, what were the bullet's flight time, its maximum height reached, and the rifle's angle of elevation?

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Solution

Several readers grumbled that this was too easy. Why not factor in air resistance? they wanted to know. And Coriolis forces, too. I'm surprised no-one wanted to bring in the Moon's gravitational influence.

Come on. I'm just offering readers something with which to kill five minutes of their coffee break. I don't expect you to solve differential equations. (Which you have to as soon as you factor in air resistance.)

OK, let's work it. First off, note that we are likely going to have a very small angle of elevation. Sines of very small angles are all right, but the cosines tend to begin with a long string of nines. That means that for accurate results, you need to use as many decimal places as you can in intermediate calculations.

I'll take the distance as 8120 feet, the muzzle velocity as 3071 ft/sec, and g, the acceleration due to gravity, as 32.17 ft/sec². I'll denote the angle of elevation by α, the flight time by T, and the maximum height reached h. Off we go.

•  The horizontal component of the bullet's starting velocity is 3071 cos α ft/sec.

•  If we're ignoring air resistance, there is no force to change this, so it remains constant across the entire flight distance of 8120 feet.

•  Flight time T is therefore 8120 / (3071 cos α) seconds.

•  The vertical component of the bullet's starting velocity is 3071 sin α ft/sec.

•  We can feed this into the well-known formula connecting starting velocity (u), final velocity (v), constant acceleration (a), and time (t):   v = u + at.

•  At the peak of its trajectory, the vertical velocity v is zero. The acceleration is −g. So we get

0 = 3071 sin α − 32.17 T / 2 (because the peak occurs at half the flight time).

•  Turning that around, T = 3071 (2 sin α) / 32.17.

•  Equating my two expressions for T, I get

8120 / (3071 cos α) = 3071 (2 sin α) / 32.17

•  Remembering that  2 sin α cos α = sin (2α), this yields

sin (2α) = 0.027697939177658118547040565299207

•  That means α is 0.79358900049387036492222975997847 degrees.

•  Feeding that back into the first expression for T gives a flight time of 2.6443435183807559986425061309246 seconds.

•  For the maximum height, use the standard formula relating distance (s), starting velocity (u), constant acceleration (a), and time (t):   s = ut + at 2 / 2.

•  Imagine the bullet falling from its peak to ground level (s = h, u = 0) for half the flight time (t = T / 2) under gravity (a = g). The formula delivers h = 28.118802316477311759487776271397.

•  Now we're through with the tricky sines and cosines, we can dump the extra decimal places and give our answers to the four-significant-digit accuracy the problem was stated in:

•  Answers:  2.644 seconds, 28.12 feet (or 8.571 meters), and 0.7936 degrees.