Compensating for Gravity in Archery: Finding the Correct Aim Point

[halo9909]

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

 

[mgb_phys]

Do you want to show how you got that answer?

 

[halo9909]

i got it now its 1.007m

cant exactly remmeber, keep dividing and multuiplying it and got it

 

[Hurkyl]

cant exactly remmeber, keep dividing and multuiplying it and got it

Then you didn't solve the problem.

 

[mgb_phys]

A new approach to the Monte Carlo method?

 

[Mentallic]

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 $$\Delta x=v_{x}t$$
2) Using that result, found the intial vertical velocity using $$\Delta x=v_{i}t+\frac{1}{2}at^{2}$$
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 (v_x=86,v_y=result from 2), by using $$\theta=tan^{-1}(\frac{v_{iy}}{v_{ix}})$$
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: $$h=\Delta x.tan\theta$$

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