# Motion in Two Dimensions of darts

1. Jan 31, 2010

### Keithkent09

1. The problem statement, all variables and given/known data
In the game of darts, the player stands with feet behind a line 2.36 m from a dartboard, with the bull's-eye at eye level. Suppose you lean across the line, release a dart at eye level 2.23 m from the board, and hit the bull's-eye. Find the initial velocity of the dart, if the maximum height of its trajectory is 1.45 cm above eye level

2. Relevant equations
y=v_0t+.5gt^2
x=v_0t
v_f=v_i+at
v_f^2=v_i^2+2ax

3. The attempt at a solution
I tried to find the angle at which the projectile was thrown by using inverse tangent of .0145/1.115 but that did not give me the correct angle. I tried plugging in these numbers and manipulating the equations but I could not get anything to work out.

2. Jan 31, 2010

### AEM

Here's some hints: the trajectory of the dart is a parabola. The maximum height of the dart is at the vertex of the parabola. Your initial velocity will be inclined upwards from the horizontal and you will need to use two kinematic equations.

3. Jan 31, 2010

### Keithkent09

Okay, I tried to use the two kinematics already but had no success. I only have two equations and three variables. I tried to substitute v_icostheta for the v_i in the x direction and v_isintheta for the v_i in the y direction and could not get anything to work. I am sorry I really am trying to learn this.

4. Feb 1, 2010

### AEM

Well, here are a couple more hints. The t variable is the same in both equations. The time taken to get to the top of the parabolic path is one-half the total time of flight. You should write two equations:

$$X = X_0 + V_{0x} t + \frac{1}{2} a_x t^2$$

and

$$Y = Y_0 + V_{0y} t + \frac{1}{2} g t^2$$

Take the origin at the point of release and $Y_0 = 0$ as does $X_0$.

Also $a_x = 0$.

Now eliminate t, and you should be able to find your answer. If that doesn't do it for you, post a response and I'll look at it again later today.

Last edited: Feb 1, 2010