# Projectile Motion Archery Problem

1. Sep 15, 2011

### Intex

Hey guys, i realize this question has been asked on these forums before, and i have seen those posts, but they dont make sense to me, and this question is on my test tomorrow so i really need to figure it out.

1. The problem statement, all variables and given/known data

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 61.0 m away, making a 3.00 degree angle with the ground.

So distance in the x direction is going to be 61m
and you have an angle theta that is 3 degrees

2. Relevant equations

v_f = V_i + at [eq1]
61tan3 = height that the arrow was initially fired from (i think this is where im wrong)
d = V_i * t + 0.5at^2 [eq2]
Y_f = Y_i + V_iy * t + 0.5 * a_y * t^2 [eq3]
distance = speed*time [eq4]

3. The attempt at a solution

so i got that 61tan3 = 3.197
I used this value in [eq3] as my y_i so i got
0 = 3.197 + 0 + 0.5(-9.8)t^2 and i solved for t which ended up being 0.8077s
then i plugged my time into [eq4] like so 61 = 0.877 * V and got 75.52 m/s for my velocity. But the answer to this problem is 107. What am i doing wrong?

2. Sep 15, 2011

### apelling

The angle is formed by the horizontal and final vertical velocity components not the displacement components.

V_f = V_h tan 3

Where I use V_h for the constant horizontal velocity you wish to find.

You'll need to use eq1 and eq4.

That should be a big enough hint I hope.