I came across the following physics exercise:

A high-powered 7-mm Remington magnum rifle fires a bullet with a velocity of $900$ m/s on a rifle range. Neglect air resistance.
(a) Calculate the distance $h$ such a bullet will drop at a range of $200$ m when fired horizontally.
(b) To compensate for the drop of the bullet, when the telescope sight is pointed right at the target, the barrel of the gun is aligned to be slanted slightly upward, pointed a distance $h$ above the target. The downward fall due to gravity then makes the bullet strike the target as desired. Suppose, however, such a rifle is fired uphill at a target $200$ m distance. If the upward slope of the hill is $45^\circ$, should you aim above or below the target, and by how much? What should you do when shooting on a downhill slope at $45^\circ$ below horizontal?

Part (a) was easy as $\pi$. However, regarding part (b), I'm in trouble. Intuitively, I'd imagine to aim a bit higher uphill and downhill, just like on a horizontal plane. But, according to the book, I'm wrong: I need to aim lower! Why!?


closed as off-topic by John Rennie, Kyle Kanos, Jon Custer, Qmechanic Jan 11 '17 at 17:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, Kyle Kanos, Jon Custer, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ Do the same calculation: how far will the bullet drop when it is fired 45deg up and 45deg down? Why? $\endgroup$ – DilithiumMatrix Jan 10 '17 at 20:54
  • $\begingroup$ @DilithiumMatrix -- At $\pm 45^\circ$, the bullet will drop ~$0.23$ m. So, if it drops, shouldn't I compensate by aiming a bit upwards? $\endgroup$ – Fine Man Jan 10 '17 at 21:13
  • 1
    $\begingroup$ How does that compare with the initial drop (from flat)? $\endgroup$ – DilithiumMatrix Jan 10 '17 at 23:42
  • $\begingroup$ @DilithiumMatrix -- I don't have the numbers with me but if I recall correctly it's about $0.54$. So, what is really meant is that you still have to compensate upwards, but just less than horizontally? $\endgroup$ – Fine Man Jan 11 '17 at 4:51
  • $\begingroup$ Yes - Exactly! The key is that the sight already has a built-in 'compensation' to it. $\endgroup$ – DilithiumMatrix Jan 11 '17 at 15:10

This is probably not a question which can be answered intuitively.

When shooting uphill the component of gravity normal to the incline is reduced, so for a given time of flight the bullet will fall less. However, there is now a component of gravity down the plane, which increases the time of flight. Whether these 2 effects compensate exactly, or which one dominates, is not obvious.

As DLM suggests, do the calculation again and compare the 2 scenarios.

In the extreme case of shooting directly above you (in an anticlockwise sense), the barrel will be aiming "high" (further anti-clockwise) so you would need to aim "low" (clockwise). However, we cannot presume this applies at all angles below 90 degrees, because it may channge over at some point.


The intuition for this is similar to the famous Monkey and Gun Experiment.

Suppose that a bullet is fired at a monkey, and the monkey lets go of it's branch at the same time (i.e. begins to free fall). Then the bullet will hit the monkey, because the effect of gravity on the monkey is the same as the on the bullet. They both fall the same amount. The horizontal velocity of the bullet determines the time needed to close the gap (since the two objects need to have the same x coordinate to be in collision), and thus also the fall height.

Similarly in your example, the distance $h$ is the distance that the bullet falls traveling to the target. Let's call it the compensation height.

How far is $h$, you ask? It depends on how long the bullet takes to cover the horizontal distance to the target. If the virtual target is in free fall for longer, it will have travelled further.

When we fire at the uphill target, we need to tilt upward relative to a straight shot to compensate for gravity, which reduces the horizontal component of the bullet's velocity. This means that the target falls for a longer time and so the compensation height needs to be more.

The effect is opposite downhill. We still need to tilt upward relative to a straight shot to compensate for gravity, but tilting upward when you're pointing downward already increases the bullet's horizontal velocity. So the time is less and so is the compensation height.

I made a sketch with the velocity vectors shown below, so you can get a visual feel for the effect:

bullet drop image


Not the answer you're looking for? Browse other questions tagged or ask your own question.