Kinematics In Two Dimensions - Archery

[Boyerta]

Homework Statement

Hey guys, so here's the problem I'm given:

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 65.3 m away, making a 2° angle with the ground. How fast was the arrow shot?

Homework Equations

Equations I used:

tanθ=opp/adj

Y_f = Y_i + V_yit + 1/2(a_y)t²
^For the motion in the y-direction as it's under constant acceleration in the y-direction.

V_x = (Δx/t)
^For motion in the x-direction as it has constant velocity in the x-direction.

The Attempt at a Solution

Looking at the problem, I see it wants the initial velocity, which would be the same as the initial velocity in the x-direction because the arrow is shot parallel to the ground, there is no y-component of velocity. I have a Δx (65.3 m), but no time.

I decide to use the equation I listed for the motion in the y-direction, but realize I need an initial distance in the y-direction (height).

Here's where I'm not quite sure if my method works. I draw a triangle having an angle of 2° like given in the problem and an adjacent side of 65.3 m. I use tan(2°) = opposite/65.3
From that I get an inital height of 2.28 m, which seems a bit high but reasonable.

Now I use
Y_f = Y_i + V_yit + 1/2(a_y)t²

Y_f = 0 m ; Y_i = 2.28 m ; V_yi = 0 m/s (arrow was shot parallel to the ground) ; a_y = -9.8 m/s²

-4.9t² + 2.28 = 0

t = .682 s Seems reasonable.

Now I use V_x = (Δx/t) to find the initial velocity in the x-direction.

V_x = (65.3/.682) = 95.75 m/s

However, my online homework site tells me it's incorrect. I feel that 95.75 m/s is a reasonable speed for an arrow. That's around 200 miles per hour isn't it?

I think my usage of tangent was probably incorrect but it was the only way I could make any progress on the problem.

Any hints or a push in the right direction would be greatly appreciated!
Thanks guys :)

 

[LawrenceC]

Hint:
When the arrow strikes the ground, you know the ratio of velocities. You also know that the velocity in the horizontal direction is constant and the range is that velocity multiplied by the time of flight. The time of flight can also be related to the ratio of velocities.

 

[Boyerta]

Your hint was exactly what I needed, I figured it all out and got the correct answer.
Thank you LawrenceC, I appreciate it!