Projectile motion with air resistance and ball of mass

[rfg]

Homework Statement

A ball of mass "m" is thrown vertically upward with a velocity of "vi." It experiences a force of air resistance given by F=-kv, where "k" is a positive constant. The positive direction for all vector quantities is upward. Does it take longer for the ball to rise to its maximum height or to fall from its maximum hieght back to the height from which it was thrown.

Homework Equations

I calculated the velocity as a function of time as (mg/k-vi)e^(-kt/m) = mg/k + v

The Attempt at a Solution

I believe that the projectile would take longer to fall to its initial position, because as it rises, the force of air friction and gravity work against the velocity. While falling, only air friction opposes the velocity.

 

[Erwin Kreyszig]

Homework Statement

A ball of mass "m" is thrown vertically upward with a force of "vi." It experiences a force of air resistance given by F=-kv, where "k" is a positive constant. The positive direction for all vector quantities is upward. Does it take longer for the ball to rise to its maximum height or to fall from its maximum hieght back to the height from which it was thrown.

Homework Equations

I calculated the velocity as a function of time as (mg/k-vi)e^(-kt/m) = mg/k + v

The Attempt at a Solution

I believe that the projectile would take longer to fall to its initial position, because as it rises, the force of air friction and gravity work against the velocity. While falling, only air friction opposes the velocity.

For this question you have to think about the constant deceleration, the air resistance. Would that effect the both the up and down directions. Then consisder gravity as a deacceleration on the way up, but is it not a acceleration on the way down.

Try using the equation of $$V_{final}$$= $$V_{initial} + at$$
rearrange for t, then you should see an obvious result. (initial velocity is Vi on the way up, and obviously 0 on the way down)

 

[rfg]

I apologize if I seem incompetent, but I don't entirely follow. I understand that the drag force will cause the magnitude of the acceleration to decrease throughout the motion. The initial acceleration will have a magnitude of g+kvi/m. As velocity diminishes, this value will reach g when v=0 (at the top of the trajectory). During the fall, acceleration still has a magnitude of g+kv/m, however velocity is now negative, thus as velocity increases, acceleration decreases. But why doesn't gravity work as an acceleration during the fall? Isn't it working in the same direction of the velocity?

 

[PhanthomJay]

I apologize if I seem incompetent, but I don't entirely follow. I understand that the drag force will cause the magnitude of the acceleration to decrease throughout the motion. The initial acceleration will have a magnitude of g+kvi/m. As velocity diminishes, this value will reach g when v=0 (at the top of the trajectory). During the fall, acceleration still has a magnitude of g+kv/m, however velocity is now negative, thus as velocity increases, acceleration decreases. But why doesn't gravity work as an acceleration during the fall? Isn't it working in the same direction of the velocity?

You are pretty much correct in your thinking, except that during the downward fall, the acceleration is g -kv/m downward, (gravity acts down , the air resistance acts up). During the upward journey, the acceleration is g +kv/m downward, since both gravity and the air resistance forces act dowm. Bottom line is that downward journey takes longer, as you had initially noted.

 

[sephirothrr]

Maybe I'm way off, but shouldn't the times be even?
Wouldn't the initial force imparted upon the ball change things?

On the way up, you have your force being counter-acted by gravity and air resistance.

On the way down, you have just gravity vs. air resistance.

This is just my logic here.

 

[PhanthomJay]

Maybe I'm way off, but shouldn't the times be even?
Wouldn't the initial force imparted upon the ball change things?

On the way up, you have your force being counter-acted by gravity and air resistance.

On the way down, you have just gravity vs. air resistance.

This is just my logic here.

There is an apparent error in the problem statement, as i see it; the ball has an initial velocity vi, not an initial force vi. For sure, there must be an initial force imparted to the ball by the motion of the thrower's hand, but the start point of this problem is at the point of release, wher only gravity and air resistance acts in both directions, no other forces act during the upward or downward flight. The ball decelerates non uniformly rapidly (greater than g) during the upward path, then accelerates downward non uniformly at less than g (possibly reaching a = 0 if it reaches terminal velocity) during the downward journey.

 

[sephirothrr]

There is an apparent error in the problem statement, as i see it; the ball has an initial velocity vi, not an initial force vi.

But the problem says:

A ball of mass "m" is thrown vertically upward with a force of "vi.

It is a force here.

 

[rfg]

I believe I have the problem figured out, thank you for the help all those who posted.

Btw, vi is a initial velocity, not a force. This wans a typo, sorry.

My conclusion is that the object does indeed take longer to fall. The times cannot be equal because that is not an answer choice. My reasoning is as follows:

While on the rise, the drag force works in the same direction as gravity, thus the resulting deceleration of the object is greater than the ideal 9.8m/s/s. Thus, the velocity diminishes faster and reaches zero (the top of the trajectory) faster than it would without friction. While falling the drag acts opposite gravity, thus the downward acceleration is less than the ideal 9.8m/s/s. Hence, the fall takes longer than it would without friction.