# Homework Help: Baseball physics problem

1. Sep 12, 2007

### Niles

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

1) A baseball is thrown straight up. The drag force is proportional to v^2. In terms of g, what is the y-component of the balls acc., when it's speed is half is terminal speed and it is moving up? When it is moving down?

3. The attempt at a solution

1) When the ball is moving up, the net force is: m*a = m*(-g) + v^2*D

It's the part with the half terminal speed that I am confused about.

When it is moving down, net force is: m*a = m*g - v^2*D. Again, the half terminal speed is confusing.

2. Sep 12, 2007

### andrevdh

The drag force opposes the motion of the object, so the direction is wrong in the first equation - both are negative.

As far as the terminal speed goes - the drag force increases with increasing speed. On the way up the speed decreases up to zero at the top. During the downward motion the speed increases, but only up to a point where the downward gravitational attraction is equal to the upward drag force. At this stage it will fall with no acceleration at a constant speed - the terminal speed.

On the way up the drag force is therefore assisting the effect of gravity, that is the speed of the ball is changing faster that it would with gravity alone (although it is decreasing).

On the way down the drag is opposing gravity, in effect cancelling (some) of its effect out, that is the speed of the ball is increasing at a slower rate than it would with gravity alone. This happens until the speed do not increase any more - the effect of gavity has been cancelled out completely by the drag and the ball is falling at a constant speed. From this point onwards the drag force will stay the same since the speed of the ball will not increase anymore (gravity has been "cancelled out").

Last edited: Sep 12, 2007
3. Sep 12, 2007

### Niles

Great explanation, I've printed it out, so I can have it with me from now on.

But where does the half terminal speed get in the equations?

4. Sep 12, 2007

### andrevdh

The terminal speed will take on a constant value. That is at some stage the upwards drag will be equal to the downwards gravitational attraction when the ball is on its way down. You can get this speed from your down equation ...

Your equations describes the acceleration as a function of the vertical speed of the ball. Just subs half the terminal speed into the equation to get the required accelerations.

5. Sep 12, 2007

### D H

Staff Emeritus
To elaborate a bit, suppose the ball is released from a helicopter. At first, the ball's speed increases as it falls toward the earth. This increased speed means the drag force, directed upward, increases as well. At some point, the acceleration due to drag (upward) will equal the acceleration due to gravity (downward). The net acceleration at this point is zero, meaning the velocity is now constant. This is the terminal velocity. Now suppose the ball is fired downward by a canon from the helicopter with an initial velocity that exceeds the terminal velocity. The upward drag acceleration will exceed the downward gravitational acceleration, making the ball slow down until it reaches the terminal velocity.

All you have to do to find the terminal velocity is find the velocity at which the magnitude of the drag acceleration is equal to the magnitude of the gravitational acceleration.