[Guest]

Here we are back on our flat, perfectly level rifle range at sea level, a crisp 59 degrees F., no humidity, the barometer steady at 29.92" Hg and the wind is calm. The rifle is a match .22 long rifle, which fires a 40 grain bullet at an initial muzzle velocity of 1080 feet per second. This particular bullet has a ballistic coefficient of .150 G1. The sight line of the rifle is 1.35" diectly above the centerline of the bore, is horizontal, and is 5 feet above the ground. (We will be firing from the standing position.) Two targets, each a U.S. Roosevelt dime, are placed directly in the line of sight, one at 50 feet, one at 50 yards. We know from experience that if we zero our rifle to hit the dime at 50 feet, no correction will be necessary to also hit the dime at 50 yards. We also know that a bullet fired at this zero will hit the ground at a velocity of 834 feet per second. The question is, where and when will the bullet hit the ground?

[bonanova]

Under the same conditions as the previous puzzle,

Dyaw a parabola through the muzzle and the two dimes.

See where it intersects the ground. [solve for x when y = 0.]

That's where.

Differentiate the parabola at the muzzle [get the inclination] to determine the x component of the velocity.

Divide x by it to get the time.

That's when.

A lot of red herring info here.

[Guest]

Under the same conditions as the previous puzzle,

Dyaw a parabola through the muzzle and the two dimes.

See where it intersects the ground.

[solve for x when y = 0.]

A heck of a lot of red herring info here.

lol contraire----the last puzzle was in a vacuum.

[Guest]

210 yards downrange

.673 seconds elapsed time.