# Minimum speed kinematics

1. Jun 24, 2008

### gerry73191

1. The problem statement, all variables and given/known data

A passenger is rising in a hot air balloon at a constant 2m/s. The passenger realizes she forgot her camera. A second person on the ground below the balloon throws the camera straight up when the balloon is 2.5m off the ground. What is the minimum velocity the camera can be thrown with in order to just reach the passenger?

2. Relevant equations
$$y_1(t)=2t+2.5$$
$$y_2(t)=vt-\frac{1}{2}gt^2$$

3. The attempt at a solution

Okay so I figured that at the minimum speed the camera will reach the passenger when its velocity is zero.

So figured out that its velocity is zero when $$t=\frac{v}{g}$$

okay then I set the two position equations equal getting $$2t+2.5=vt-\frac{1}{2}gt^2$$

So then my next step I wasn't sure was valid, but I substituted $$\frac{v}{g}$$ for t and then solved for v.

I got 9.28 m/s.

I don't know if this answer is correct or not.

Can someone please confirm my methods?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 24, 2008

### calef

You got it.

Question: Is this problem for a course involving calculus?

3. Jun 24, 2008

### gerry73191

no its for AP Physics B which is algebra based.

I know Calculus, well two semesters worth. How would I solve it using calc?

4. Jun 24, 2008

### calef

I just noticed that the derivative of your y2(t) equation effectively yields your t value, after a little majiggering.

Velocity is the derivative of displacement, therefore, when velocity is zero, your displacement graph will either be a peak or a trough. So when you take the derivative, you just set it equal to zero, and isolate t.

Which is what you did, actually, but I wasn't sure if you knew that's what you were doing, or if you were just using an equation (all the equations are derived from calculus, anyway).