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I've read from different sources (can't find them online) that firing a bullet at different angles from horizontal will result in differing amount of bullet drop.

For example, a gun is fired at a target 2000 yards away horizontally. The bullet drops X inches. Again the gun is fired at a target 2000 yards away with a 30 degree angle upward. The bullet drops Y inches. X does not equal Y. Same scenario, only 30 degrees downward angle. Bullet drops Z. Z does not equal X or Y. Why? What is happening? The bullet is fired at the same speed at the same distance, the bullet should be in the air for the same time. Acceleration due to gravity is a function of time, so the bullet should drop equally in all cases. Right?

The only change in each case is the angle gravity acts on the bullet. My guess is that when the bullet is fired upwards, gravity is pulling it downward, thus slowing the bullet down causing it to be in the air longer and increasing the drop.

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Assuming you are ignoring air friction, then if $v_0$ is the muzzle speed of the bullet and $\theta$ is the angle from the horizontal, then the horizontal speed will be $v_h = v_0 cos(\theta)$ and the vertical speed will be $v_v = v_0 sin(\theta)$. So the bullet will reach a horizontal distance of $d = 2000 yards$ at a time $t = d/v_h$, so the time of flight will depend on the angle but the time for $+30^o$ and $-30^o$ will be equal and longer than the $0^o$ case. The distance that the bullet will fall relative to the horizontal axis, $\Delta y$, will be given by $\Delta y = v_v t -\frac{1}{2} g t^2$. So it should be clear that all three $\Delta y$'s will be different since $v_v$ has different signs for $+30^o$ and $-30^o$.

However, if you are measuring the drop relative to the aim point, which is at $v_v t$ then the drop will be $\Delta y = -\frac{1}{2} g t^2$. Now the $\Delta y$ for the $+30^o$ and $-30^o$ will be equal and different than the $0^o$ case.

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  • $\begingroup$ Are you assuming that the target is 3000 yards away horizontally in all cases? What about gun and target are 3000 yards away regardless of angle? So then would t=d/v0? And time to reach target would be equal in all cases? How is drop relative to the aim point differ than the drop in the first paragraph? Isn't it the same thing? $\endgroup$
    – SaulBack
    Oct 21, 2011 at 13:33
  • $\begingroup$ What would adding air friction do to the equation? $\endgroup$
    – SaulBack
    Oct 21, 2011 at 13:54
  • $\begingroup$ Yes, if the target is 2000 yards along the aim direction then we would have $t = d/v_0$ and it would be the same in all 3 cases. Then the drop from the aim point would also be the same in all 3 cases. My first paragraph is calculating the y position of the bullet when it gets to the 2000 yard distance. The second gives the drop from the point where the bullet would have hit if there was no gravity. Hope that is clear. Air friction would make this much more complicated and would depend on a model for the air friction force (usually it is quadratic in air speed). Ask a new question... $\endgroup$
    – FrankH
    Oct 21, 2011 at 18:32
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The horizontal component of the initial speed is irrelevant. The vertical component matters only. So answer this question how long does it take for an item to drop to the ground after it's been toss upwards with $v_y$ speed (starting from a height $h$)?

The answer comes from $y = h + v_y t + \frac{1}{2} g t^2 = 0 $ with solution $$ t = \frac{\sqrt{2 g h + v_y^2}+v_y}{g}$$

So as other posts have pointed out, as the angle of the shot changes the vertical velocity changes, and that is what determines the time of flight.

The horizontal component only affects the range of the the shot, not the time!

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  • $\begingroup$ You are misunderstanding the statement of the problem. He is not firing the shot and seeing where it hits the ground. He is aiming at a target that is 2000 yards away horizontally, so the horizontal speed will determine how long the bullet is in the air before it hits the target. $\endgroup$
    – FrankH
    Oct 21, 2011 at 6:21
  • $\begingroup$ Further, the air friction is essential. You can't answer real-world problems with textbook crap. $\endgroup$
    – Ron Maimon
    Oct 21, 2011 at 20:53
  • $\begingroup$ Then how are you going to hit the target if you vary the angle of departure? In real life, you don't choose the exit velocity of a bullet. $\endgroup$
    – John Alexiou
    Oct 21, 2011 at 23:16
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I realize this thread is three years old. In real life, you measure the angle of trajectory with MOA. Or minute of angle, rifle scopes mostly come standard with these MOA adjustments. Minute of angle adjustments equate to 1/60th of each degree out of 360 degrees. On average one MOA is ~1 in. of bullet drop per 100 yds. Out to 2000 yards you will most likely be spending a small fortune on getting the rifle and scope set up. (Mounting scope on a decline, kind of pointing towards the barrel to compensate for the extreme bullet drop) Take into account velocity, twist rate of barrel, ballistic coefficient of projectile, drag coefficient, weight of projectile, air density, temperature, and multiple varying wind magnitudes and you have yourself a fun physics problem. :D

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it seems you are wrong take a look at this http://www.backcountrymaven.com/journal/2013/5/16/understanding-uphill-and-downhill-shots-in-long-range-shooti.html

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