Projectile Motion Archer Problem

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The discussion focuses on solving two-dimensional projectile motion problems involving an archer. The first problem involves calculating the maximum height of an arrow fired at an angle, which can be simplified by decomposing the initial velocity into vertical and horizontal components. The second problem requires determining the initial velocity needed for an arrow to hit a target at a lower height, emphasizing the importance of vertical motion due to gravity. Key strategies include treating the horizontal and vertical motions as separate one-dimensional problems. The explanation clarifies how to approach these types of projectile motion challenges effectively.
nickelodeon77
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I need help with a certain type of projectile motion problem (dunno exactly what it's called), but here's an example (these problems don't include air resistance):

An archer lies on the ground and fires an arrow with an initial velocity of 112 ft/sec at 40 degrees. What is the maximum height that the arrow will reach?

Another problem that I have trouble with is the type of projectile motion that involves something going from a certain height to a lower height. Here's an example of this:

An archer stands on a hill so that the arrow in the bow is 20 m above the groundn and is aimed at an angle of 0 degrees relative to the ground. If the archer is trying to hit a target that is lying on the ground 118 m away, with what initial velocity must she launch her arrow?

Can anyone explain how to solve these problems? Thanks for the help

-Nick
 
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Both of these problems are two-dimensional kinematics problems. Do you understand how to solve one-dimensional problems? For instance, if the archer had fired the arrow straight up instead of at an angle, could you solve that one?
 
Yeah, I know how to solve one-dimensional problems. I just can't seem to figure these out though.

-Nick
 
OK. The key here is to realize that you can decompose two- (or more-) dimensional problems into that many coupled one-dimensional problems. For instance, consider your "archer" question. The height the arrow reaches is really only determined by things that act vertically - gravity and the initial vertical speed of the arrow. If you decompose the initial speed of the arrow into horizontal and vertical components, then you can ignore the horizontal part and just work vertically. Voila! It's a one-dimensional problem.

Some students have a hard time seeing that the horizontal component of the velocity is irrelevant in that case. Does that make sense to you?

In the second case, it's a bit more complicated. You're looking for the range of the arrow. That's a "horizontal" question. However, it will travel horizontally only so long as it's in the air, and that's a "vertical" question. So: you'll split the velocity into horizontal and vertical components again. Realize that the only acceleration is vertically, so the horizontal side of things is just a "rate times time equals distance" problem - all you're missing is the time. Use the vertical parts to find the time, plug it into the horizontal part, and Bob's your uncle. In this case, the problem is a little easier than it could be because the initial vertical velocity is 0, but you can't ignore the vertical since that's the way gravity acts.

Does that help?
 
Yeah! A whole lot actually. Thanks!

-Nick
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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