Fired Bullet Formula: Calculate Fallout Height

[Zipher]

ok here's the thing. i fire a bullet at velocity (let's say static velocity, no acceleration) of 500m per sec. let's say that bullet's mass is 5g. what is the formula of its fallout?
i mean if i fire the bullet at height 2m straight forward, at what height would it be at 1000m after fired?

 

[Mech_Engineer]

ok here's the thing. i fire a bullet at velocity (let's say static velocity, no acceleration) of 500m per sec. let's say that bullet's mass is 5g. what is the formula of its fallout?
i mean if i fire the bullet at height 2m straight forward, at what height would it be at 1000m after fired?

If you assume that is is not decelerating in air (assume it's flying in a vacuum) then it will only accelerate towards the ground. You can find how long it would take it to fly 1km, and then calculate how far it fell in that amount of time. Use your kinematic equations, it's essentially 1/2 of a ballistic trajectory. It sounds to me like the bullet will need to be going faster or it might hit the ground before it reaches 1km, however.

EDIT: By the way, if the bullet is traveling in a vacuum, its mass does not matter. Interesting side note...

 

[brewnog]

Theoretically, the bullet will fall at exactly the same rate as it would if you just dropped it. Its vertical acceleration due to gravity is independent of its horizontal motion.

 

[nrqed]

ok here's the thing. i fire a bullet at velocity (let's say static velocity, no acceleration) of 500m per sec. let's say that bullet's mass is 5g. what is the formula of its fallout?
i mean if i fire the bullet at height 2m straight forward, at what height would it be at 1000m after fired?

$$y_f = y_i + v_{yi} t - {1 \over 2} g t^2$$
Setting the origin at the initial point and assuming that the initial fdirection is purely horizontal, this simplifies to $y_f = - {1\over2} g t^2$.

In the x direction (again setting the origin at the initial position)
$$x_f = x_i + v_{x} t = v_{x} t$$

Isolating t in the second formula and plugging in the first, one gets
$$y_f = -{1 \over 2} g { x_f^2 \over v_x^2 }$$

Which shows the parabolic shape of the trajectory and its curvature. In that form, you simply have to plug in the value of the horizontal distance traveled and you get the value of the y coordinate.

Of course, this has nothing to do with real trajectories because of the large effect of air drag. (and if v_x is is made huge, one would have to worry about the curvature of th Earth at some point )

(and if the initial velocity is really huge, at some point one would have to take into account that the force of gravity is not simply given by mg)

 

[Integral]

As Brewnog said, it will take the same amount of time to reach the ground as if it fell from 2m (~.6 s) so at a velocity of 500m/s the bullet will have a range of about 300m.

Just apply the basic equations of a falling body.

$$s = -\frac 1 2 g t^2 + V_0 t + h$$

to get the time of flight, then use that time and the constant horizontal velocity to get the range.