Projectile Motion Archer Problem

In summary, the archer will reach a maximum height of 118 m if she launches her arrow with an initial velocity of 112 ft/sec.
  • #1
nickelodeon77
3
0
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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  • #2
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?
 
  • #3
Yeah, I know how to solve one-dimensional problems. I just can't seem to figure these out though.

-Nick
 
  • #4
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?
 
  • #5
Yeah! A whole lot actually. Thanks!

-Nick
 

1. What is projectile motion in an Archer Problem?

Projectile motion in an Archer Problem refers to the path that an arrow takes when it is fired from a moving archer. It is the combination of two motions - horizontal motion and vertical motion.

2. What factors affect projectile motion in an Archer Problem?

The factors that affect projectile motion in an Archer Problem are initial velocity of the arrow, angle of release, air resistance, and gravity. These factors can change the trajectory and distance of the arrow.

3. How do you calculate the range of a projectile in an Archer Problem?

The range of a projectile in an Archer Problem can be calculated using the formula: R = (V^2 * sin(2θ)) / g, where R is the range, V is the initial velocity, θ is the angle of release, and g is the acceleration due to gravity.

4. How does air resistance affect the trajectory of a projectile in an Archer Problem?

Air resistance can affect the trajectory of a projectile in an Archer Problem by slowing down the horizontal velocity of the arrow. This can result in a shorter range and a steeper trajectory.

5. What is the maximum height reached by a projectile in an Archer Problem?

The maximum height reached by a projectile in an Archer Problem can be calculated using the formula: H = (V^2 * sin^2(θ)) / 2g, where H is the maximum height, V is the initial velocity, θ is the angle of release, and g is the acceleration due to gravity. This is the highest point on the projectile's trajectory before it starts to fall back to the ground.

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