# Terminal speed and acceleration problem

1. Oct 9, 2009

### MTBrider

I'm having trouble figuring this one out. Hope someone can help

A ball dropped from rest accelerates at a rate of 9.81 m/s2 if we ignore air resistance. If air
resistance is not ignored, the ball reaches a terminal speed when the friction force of the air
acting on the falling ball is equal to the gravitational force acting upon the ball. If, instead of
being dropped, the ball is hurled upwards into the air at a speed greater than the terminal
speed the acceleration of the ball is…
a) 9.81 m/s2
b) less than 9.81 m/s2
c) greater than 9.81 m/s2 but less than 19.62 m/s2
d) 19.62 m/s2 or greater

I'm thinking you would need more information to actually pin down a number for the acceleration but I may be wrong. Would it be C, greater than g but less than 2g?

Last edited: Oct 9, 2009
2. Oct 9, 2009

### willem2

At greater than terminal speed the acceleration due to air friction is (less/greater/equal) than g.

this acceleration is in the (same/opposite) direction as the acceleration due to gravity.

the sum of both is the acceleration you want.

3. Oct 9, 2009

### MTBrider

Thank you, got it worked out

4. Oct 9, 2009

### Franco_v

When the ball is falling downward at terminal speed the frictional air resistance is acting opposite to the direction of gravity, and the ball doesn't accelerate (since it's moving at constant velocity). Therefore, Friction = mg.

Air resistance/friction increases as speed increases. So if the ball is thrown up at a speed greater than the terminal speed, the force of air friction is at least equal to mg.

When the ball is thrown up, gravity and air resistance both act in the same direction (down).

Therefore, the minimum force acting on the ball is the force due to air resistance PLUS the force due to gravity, which is F = mg + mg = 2mg

Since F = ma, 2mg = ma, and a = 2g = 2*9.81 = 19.62 m/s^2 (minimum)

The acceleration of the ball is d) 19.62 m/s^2 or greater
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