# Very simple drag question

1. Feb 11, 2008

### Monocles

[SOLVED] very simple drag question

1. The problem statement, all variables and given/known data
A ping pong ball of mass m is thrown downward at twice it's terminal speed. What is its magnitude of acceleration immediately after release in terms of m and any other physical or mathematical constants?

2. Relevant equations
This was a quiz question, the only ones I could think of were
D = 1/4Av^2
Sum of Forces = ma = D - mg

3. The attempt at a solution

Since we weren't given a drag coefficient or area of the ping pong ball I had no idea how to proceed. All I knew is that we were allowed to use the ball's mass and any other physical or mathematical constants to solve it. Since mass was the only given quantity allowed I figured that it wasn't involved in the solution because mass isn't a dimension of acceleration. I couldn't find the drag force on the ball, so I just wrote that the magnitude of acceleration was g. It was pretty much a pure guess. Maybe this is the correct answer, but this is only question on the quiz I had no idea how to do so I wanted to ask here as soon as possible :)

2. Feb 11, 2008

### PhanthomJay

assuming quadratic drag, the drag force is
F_drag = kV^2. Calculate the terminal velocity, then the drag force based on the speed at release, then use your correct 'sum of forces' equation. Most stuff should cancel. The acceleration would initially be 'g' only if the object were dropped from rest. That's not the case here.

3. Feb 11, 2008

### Monocles

Alright I think I have it. The equation we were told to use for drag was 1/4Av^2, so from that:

Fdrag = drag force
Vt = terminal velocity
A = area

Fdrag @ Vt = mg = 1/4AVt^2
Fdrag @ 2Vt = 1/4A(2Vt)^2 = AVt^2
Vt = sqrt(4mg/A)
ma = AVt^2 - mg
=A(4mg/A) - mg
=4mg - mg

therefore magnitude of acceleration = 3g

4. Feb 11, 2008

### lavalamp

At terminal velocity, you know the following:

Force due to gravity = force due to drag
$$mg = 0.5 * \rho * v^2 * AC_{d}$$

At twice terminal velocity you know that the force due to drag is:
$$F_{d} = 0.5 * \rho * (2v)^2 * AC_{d}$$
$$F_{d} = 0.5 * \rho * 4 * v^2 * AC_{d}$$

That's four times as high as before, since drag increases with the square of the velocity. Obviously mg hasn't changed though, so you know that overall the total force acting on the ball is 3mg. Therefore the acceleration on the ball is 3g vertially upwards:
$$3 * 9.8 = 29.4 m/s^{2}$$