# Water Balloon Launcher

1. Jan 12, 2009

### manslayer68

1. The problem statement, all variables and given/known data
Given
hb (m) = 1 (Height of point B)
theta = 30 (degrees - launch angle)
k (N/m)= 60 (assume the graph is linear)
m (g) = 250 (balloon mass)

1. If Xs (cm) = 150 (pull distance)
a) Find the estimated Vb and the landing X (ideal case where Ein =Eout)
b) The balloon land 160 ft away. Find the actual Vb for the shot.
Find also the Eout and the efficiency of the launcher.

2. Relevant equations
1/2kx^2
mgh
1/2mv^2

3. The attempt at a solution

dont know how to solve for a)...not sure with the what i am doing. any hint on what formula i should be using?
for b, i put ft into m, so x = 48.7656m and used it to find time, t = 2.439s, which is then use to find Vb, Vb = 23.087m/s

so how exactly do i solve for a) and find Eout and the efficiency ?

Last edited: Jan 12, 2009
2. Jan 12, 2009

### Delphi51

It looks like you have a spring or elastic launcher with a spring constant of k.
Is B the launch point? Vb the launch velocity?
If so, (a) involves finding the speed of the balloon when it leaves the launcher.
You are directed to use conservation of energy. That is, the energy of the spring is converted entirely into the energy of the moving balloon. Start with
SPRING ENERGY = KINETIC ENERGY
and put in the formulas for those two types of energy.

3. Jan 13, 2009

### manslayer68

yes Hb = Yo and Vb is the launch velocity

k for Ein=Eout, i did 1/2kxs^2=1/2kmv^2 and got an efficiency of 98.7%

Vb ideal is found by V = Squareroot((KXs^2)/m) where Vb is 23.24m/s

but i still cant find the X ideal

here what i gotten for the approach to X: X=Vo/Cos(theta)t but cant find X if i dont know t(time)
X=1/2at^2 + Vbt and using this approach still doesnt help...

Last edited: Jan 13, 2009