(adsbygoogle = window.adsbygoogle || []).push({}); 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 = v_{o}+at

x = x_{o}+v_{o}t+(1/2)at^{2}

v^{2 }= v_{o}^{2}+2a(x-x_{o})

Range = v_{0}^{2}sin2θ/g

3. The attempt at a solution

So, I know that my distance is 50 yds. I know my angle is 45^{o}, because 45^{o}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 V_{o}Sin/Cos(theta)= V_{oy}/_{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 = v_{0y}^{2}sin2θ/g

Range = 50 yd

v_{0y}= ?

θ = 45

Ah, so that may work...

So, v_{0y}=Square root((Range*g)/(Sin2θ))

So, v_{0y}=Square root(( 45.72m*9.81m/s^2)/(Sin2(45))

So, v_{0y}=22.40 m/s = 24.49 yd/s

Okay, so that's the initial velocity in the y direction. Progress.

So, V_{o}Sin(theta)= V_{oy}Then:

V_{o}=?

theta=45

V_{oy}=22.40 m/s

So then, V_{o}=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 v^{2}= v_{o}^{2}+2a(s-s_{o}), 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Calculating the trajectory of a projectile

**Physics Forums | Science Articles, Homework Help, Discussion**