# Finding release angle and final velocity of projectile when the target has a different coordinations [closed]

I am trying to write some codes for auto-snowball mechanism which seems like a regular projectile but instead of hitting the targets on the ground it is going to be able to hit the target in the air or with different(x,y) coordination.

Here is the blueprint of the mechanism: :

Simply The mechanism will be able to get the (x,y) of the target and then mechanism will have a constant initial velocity(Vo).

I used regular Velocity equations to find the Vox and Voy of Initial Velocity(Vo).

Here is my Velocity and Delta X and Delta Y formulas:

My questions are:

1-Is a(acceleration) still going to be a=0 for Delta X and a=-g=-9.8 m/s^2

2-How am I going to find the final Velocity(Vf)?

3-Is this a correct equation to find Release Angle(Theta)?

• Your equations are somewhat unreadable. I suspect you'll get better responses if you use the MathJaX/LaTeX functionality of this site. Also, why do you want the final velocities?
– BMS
Mar 29 '14 at 19:51
• Mar 29 '14 at 20:08
• BMS-The reason I would like to find the Vf(Final velocity) is to print with what velocity it will hit the target.Nothing really special. Mar 29 '14 at 20:34
• The title should be "Design the release angle.." as you want to find the desired angle to hit a target, instead of calculating the actual release angle from current geometry. Mar 30 '14 at 19:15

$$\Delta y = v_{0}\sin{\theta} \Delta t - \frac{1}{2}g\Delta t ^2 \\ \Delta x = v_{0}\cos{\theta}\Delta t$$
You can rearrange to eliminate $\Delta t$:
$$\Delta t = \frac{\Delta x}{v_0 \cos{\theta}}$$
Substituting this into the $\Delta y$ equation yields:
$$\Delta y = \Delta x \tan{\theta} - \frac{g(\Delta x)^2}{2v_0^2\cos^2{ \theta}}$$
Now we have an equation independent of time, and it describes the parabolic path taken by the particle. Solving this equation for $\theta$, given your target coordinates $(x,y)$ will tell you the angle to fire at. Note that there will be two angles that will hit the target, unless $45^\circ{}$ is the exact answer.