Archer's Aim: Finding the Speed of an Arrow

[wadini]

An archer shoots an arrow horizontally at a target 17 away. The arrow is aimed directly at the center of the target, but it hits 51 lower.

What was the initial speed of the arrow?

so I know I need to find time and then solve for the distance

so I entered in 17^2 + .51^2 and got 17.0076^2 and tried to solve for distance and time but that did not work...I really have no idea what I am doing. How do you work these types of problems?

 

[LowlyPion]

An archer shoots an arrow horizontally at a target 17 away. The arrow is aimed directly at the center of the target, but it hits 51 lower.

What was the initial speed of the arrow?

so I know I need to find time and then solve for the distance

so I entered in 17^2 + .51^2 and got 17.0076^2 and tried to solve for distance and time but that did not work...I really have no idea what I am doing. How do you work these types of problems?

What time interval is involved with it dropping 51 cm ?

Armed with the time and distance I'd guess you know how to do the rest.

 

[tomotron3000]

Dude, you need units. 17 meters,feet? hits 51 lower inches cm?
I would say, why does the arrow hit 51(?) lower?
gravity?
How long does it take for gravity to move something 51(?)?
Then it would take the same time for the arrow to reach the tree, then for the arrow to fall 51(?).
Hope this helps. I needed help myself on another problem, and while waiting for a response, I have decided to try and help others.
Good luck!

 

[Pengwuino]

Well, for one, remember your units! I'm assuming it was 17 meters away but hit 51 cm lower?

The one definite thing we know is that the arrow fell 0.51m in the distance it traveled because we KNOW what free fall is like by our kinematic equations. Remember, vertical travel and horizontal travel are independent. Since there is no initial velocity and the arrow is under a constant vertical acceleration, the kinematic equation is $$y = y_0 + v_{0y} t + \frac{{gt^2 }}{2}$$. Let's assume that our origin is where the arrow starts at so that $$y_0 = 0$$. Also we know there is no initial velocity in the vertical direction so taht $$v_{0y} = 0$$. Using this, you can immediately calculate the time. Remember, the signs can get confusing so if you stick with g = -9.8m/s^2, the position will be a -0.51m.

Now you can determine the horizontal velocity (and in this case, since there is no angle, the initial speed) with $$x = x_0 + v_{0x} t$$. Again, our arrow starts at the origin so $$x_0 = 0$$. You can immediately solve for $$v_{0x}$$ now.

 

[wadini]

im sorry the 17 is meters and the 51 is cm

 

[wadini]

Thank you Pengwuino! That made perfect sense? Is there anyplace I can get a list of equations? I am working without a book so I keep having to find equations online. Thank you!

 

[Pengwuino]

The sticky in this sub-forum has a great list of equations.

https://www.physicsforums.com/showthread.php?t=110015