Very well, but I wonder; if a pitcher pitches a a fastball at 98mph parallel and at the same time - someone drops a ball at the same height; I'm inclined to believe the dropped ball will hit the ground first.

 

I wasn't aware that answering the question was actually relevant to anything important. I'll raise my hand and wait my turn next time.

 

but isn't the weight of the dropped unfired bullet greater because it still has the casing, powder, and whatnot? I mean that's getting into the nitty gritty and I'm not sure if that was intended to be included in the answer.

 

Clarification needs to be made. Does 'unfired bullet' simply mean the bullet or does he mean an actual complete round?

 

weight does not affect the acceleration of an object due to the pull of gravity.


also, terminal velocity is a function of the object meeting its maximum speed due to the medium it is traveling through (water, air, sand...)

you will not reach terminal velocity from a 5ft drop, but it will be something that is encountered at much greater distances.

 

According to the problem, the ground is perfectly flat. That gets rid of curvature of the earth idea. The acceleration due to gravity is 9.8 m/s^2 regardless of how fast something is moving. Discounting wind, friction, or any other parameter not specifically stated in the problem, they will, according to the law of gravity, hit the ground simulaneously.

"centerline of the bore" means that the exact middle of the projectile will be 36" off the ground.

The weight of the round, fired or unfired, shouldn't make a difference either, as there is not enough distance (only 3 feet) to reach terminal velocity anyway.

Besides, without a constant of friction for the fired and unfired projectiles against the air itself, there is not enough from which to draw a conclusion as to which will fall faster.

The answer is, the same time without more data.

 

but gravity takes place of the earths roundness....

so i would say that the shot bullet will hit first due to gravitational pull towards earth. it may be shot at a parallel angle to the ground, but as soon as it leaves the barrel of the gun gravity is pushing down on it

 

ah, got it, so unless the drag of the two objects is greatly different, they will pretty much hit the ground at the same time (the difference from slightly different drag would be negligible).

I haven't taken a physics class since sophomore year of high school

 

They hit at the same point in time. Gravity acts equally on both bullets, regardless of horizontal speed (ie being fired or not). Mass plays no role in their speed to the ground.

Converting 36" to meters, I find that the dropped bullet reaches a final velocity of 4.223 m/s the instant before touching the ground. As such, it takes 0.431 seconds to hit. Since the same gravitation force applies to the fired bullet, we now know that it will be airborne for the same amount of time, so at 3400 f/s, the bullet travels 1465 feet away from the gun before touching down (constant horizontal velocity strictly assumes no air drag).

Now, as to the question of the curvature of the earth. While valid at higher velocities (like orbital or sub orbital velocities in the 20,000 mph range), it is not the case here. The circumference of the earth is something like 131 million feet (if my units are right), which means the angle subtended by the traveling bullet along the surface arc of the earth (in 0.431s) is a measly 0.000012 degrees. This is also known as 'jack sh$t'. By use of the small angle approximation, the curvature of the earth plays no meaningful role in this problem.

 

beacuse of the spin he puts on i t

air holds it up

 

Is that the scientific definition and word for it?