Compensating for Gravity in Archery: Finding the Correct Aim Point

In summary, the problem involves an archer standing 39.0 m from a target and shooting an arrow horizontally with a velocity of 86.0 m/s. The question is asking how far above the bull's-eye the archer must aim to compensate for gravity pulling the arrow downward. Using the equation d=vit + at^2, the approximate answer is 1.007m. However, one person also used a more complicated approach involving trigonometry and velocity vectors to find the angle of the arrow and its distance from the target.
  • #1
halo9909
37
0

Homework Statement




An archer stands 39.0 m from the target. If the arrow is shot horizontally with a velocity of 86.0 m/s, how far above the bull's-eye must he aim to compensate for gravity pulling his arrow downward?

Homework Equations



d=vit + at^2

The Attempt at a Solution



I ended up getting 2.01m but am sure if that is correct and would like to know if I am not how to go about this, since it is confusing, when you don't know what formulas should/can be used
 
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  • #2
Do you want to show how you got that answer?
 
  • #3
i got it now its 1.007m

cant exactly remmeber, keep dividing and multuiplying it and got it
 
  • #4
halo9909 said:
cant exactly remmeber, keep dividing and multuiplying it and got it
Then you didn't solve the problem.
 
  • #5
A new approach to the Monte Carlo method?
 
  • #6
wow that works now? I wish I had that talent

I also got the same result as you, but my method was a little more complicated than just "multiplying and dividing" until I got the answer. How about we share our approaches so as we can learn a little something off each other?

Mine:

1) I calculated the time of flight using [tex]\Delta x=v_{x}t[/tex]
2) Using that result, found the intial vertical velocity using [tex]\Delta x=v_{i}t+\frac{1}{2}at^{2}[/tex]
3) I was then able to find the acute angle that the arrow had to make with the horizontal by trigonometry and the velocity vectors (vx=86,vy=result from 2), by using [tex]\theta=tan^{-1}(\frac{v_{iy}}{v_{ix}})[/tex]
4) Once I found the angle of the arrow to the horizontal, I could find exactly how far above the bullseye the arrow was pointing at by considering the distance between the arrow and target and using trigonometry: [tex]h=\Delta x.tan\theta[/tex]

From the sound of it, you have taken another approach which sounds more simple. If possible, could you please share your solution?
 

What is the displacement of an archer?

The displacement of an archer refers to the distance and direction between the starting point and ending point of their movement or shot. It can be measured in units such as meters or feet.

What factors affect the displacement of an archer?

The displacement of an archer can be affected by several factors, including the angle of the shot, the force applied to the arrow, the weight of the arrow, and external forces such as wind and gravity.

How is the displacement of an archer calculated?

The displacement of an archer can be calculated by using the formula: displacement = initial velocity * time + (1/2) * acceleration * time^2. The initial velocity is the speed at which the archer releases the arrow, and the acceleration is the rate at which the arrow's speed changes over time.

Why is the displacement of an archer important?

The displacement of an archer is important because it determines the distance and direction of their shot, which can greatly impact their success in hitting a target. It also allows for the calculation of other important factors such as velocity and acceleration.

How can the displacement of an archer be maximized?

The displacement of an archer can be maximized by adjusting various factors such as the angle of the shot, the force applied to the arrow, and the weight of the arrow. Additionally, proper technique and practice can also help improve an archer's displacement and accuracy.

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