# Homework Help: Calculating the trajectory of a projectile

1. Sep 27, 2010

### ://Justice

1. The problem statement, all variables and given/known data
So, we have a physics project, in which we must build a water balloon launcher and launch the balloon at a stationary target about 50 yds away (normally we use meter, however we are firing on a football field). I have created my launcher (it is basically a slingshot), but now I must find certain aspects of the trajectory. I must find:
(a) Initial Velocity
(b) Maximum height
(c) Angle of Launch

2. Relevant equations
v = vo+at
x = xo+vot+(1/2)at2
v2 = vo2+2a(x-xo)
Range = v02sin2θ/g

3. The attempt at a solution
So, I know that my distance is 50 yds. I know my angle is 45o, because 45o is the degree at which one gets the most effective range. Now, for my time, I have not yet actually calculated it, so lets just say it's... 1.78 s (does that sound reasonable?). I am pretty sure that given this information, it shouldn't be too difficult for me to find my unknowns.
I'll attempt to start by finding the initial velocity... Now, initial velocity, I know, is quite a bit different from my average velocity (total distance/total time). So... I know that in the vertical direction, the velocity is affected by gravity, and (disregarding air resistance and such), there is no positive or negative velocity in the horizontal direction. I do have my angle. So, there is VoSin/Cos(theta)= Voy/ox. Umm.... now I am a bit stuck...
Once I find the initial velocity, finding the maximum shouldn't be too hard, however I still may need a bit of help with it...

Or, perhaps Range = v0y2sin2θ/g
Range = 50 yd
v0y = ?
θ = 45
Ah, so that may work...
So, v0y=Square root((Range*g)/(Sin2θ))
So, v0y=Square root(( 45.72m*9.81m/s^2)/(Sin2(45))
So, v0y=22.40 m/s = 24.49 yd/s
Okay, so that's the initial velocity in the y direction. Progress.

So, VoSin(theta)= Voy Then:
Vo=?
theta=45
Voy=22.40 m/s
So then, Vo=26.32m/s. Awesome, found it. Still would like to make sure it's correct, so I will continue to post this.

So, now maximum height. I think I will go with v2= vo2+2a(s-so), using the y component.
v=0 (since the velocity in the y direction at max height is 0 )
v0y=22.40 m/s
a=-g
s=?
s0=0
SO, set up to solve for x... So, (-Vo^2)/(-2g). So, x=25.57m
Is that correct? If so, awesome. Thanks for all the help! haha.

2. Sep 27, 2010

### ://Justice

*bump* so could someone just confirm that I did this correctly? Please? This is due tomorrow....