# Velocity in relation to position

1. Feb 17, 2013

### destinc

A hot-air balloon has just lifted off and is rising at the constant rate of 1.8 . Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward. If the passenger is 3.1 above her friend when the camera is tossed, what is the minimum initial speed of the camera if it is to just reach the passenger? (Hint: When the camera is thrown with its minimum speed, its speed on reaching the passenger is the same as the speed of the passenger.)

I tried to use V^2=Vi^2 + 2aΔX to solve for Vi
I plugged in (1.8)^2=Vi^2 +2(-9.8)3.1
and got Vi=8m/s but that answer is wrong. Where did I go wrong?

2. Feb 17, 2013

### SammyS

Staff Emeritus
First of all, you need some units on the given data.

Then, "rising at the constant rate of 1.8" could mean a constant velocity, or a constant acceleration. Again, units would help clear up this ambiguity.

3. Feb 17, 2013

### destinc

sorry, balloon is rising at constant rate of 1.8m/s.
balloon height is 3.1m.
for the formula I used (1.8m/s)^2=Vi + 2(-9.8m/s^2)3.1m

4. Feb 17, 2013

### SammyS

Staff Emeritus
You are mixing the velocity and altitude of the balloon with the acceleration of the camera. Also vi should be squared in this formula. Beside that, I doubt that this kinematic equation is the most helpful in solving this problem.

Express the altitude of the balloon as a function of time, t, where t = 0 seconds at the moment at which the balloon's altitude was 3.1 m, which is also the moment the camera was tossed up .

Express the altitude of the camera as a function of time, t .

Equate the two & solve for t. The result will depend on the initial velocity of the camera.