Which acceleration is faster: Throwing a ball or falling due to gravity?

  • Thread starter Thread starter nhmllr
  • Start date Start date
  • Tags Tags
    Acceleration
AI Thread Summary
The discussion centers on comparing the acceleration of a baseball while being thrown versus after it leaves the thrower's hand. It is established that the acceleration due to gravity after release is 9.8 m/s², while the acceleration during the throw depends on the force applied and the time taken to reach the ball's release velocity. Participants emphasize that the time of acceleration during the throw is much shorter than the time taken for the ball to decelerate under gravity until it reaches maximum height. The conversation highlights the importance of understanding the relationship between acceleration, velocity, and time in this context. Ultimately, the acceleration experienced while throwing the ball is greater than the acceleration due to gravity once it is in free fall.
nhmllr
Messages
183
Reaction score
1

Homework Statement


You throw a baseball straight up in the air so it rises to a maximum height much greater than your height. Is the magnitude of acceleration greater while it is being thrown or after it leaves your hand. Explain.

Homework Equations


acceleration = velocity/time


The Attempt at a Solution


I know this is a really silly problem, but I just don't know the answer to it. The magnitude of the acceleration of the ball after it's left your hand would be due to gravity, which is 9.8 m/s^2. But is that more or less than the acceleration than the ball experiences while you throw it? Doesn't that completely depend on the amount of time it takes you from when your throwing arm is at rest to when it has a velocity and let's go of the ball?

But then they say that maximum height is "much" greater than my height. But doesn't that have to do with the velocity of the ball when it's left your hand, not the acceleration it experienced while you threw it?

I'm probably over-thinking this. Thanks
 
Physics news on Phys.org
nhmllr said:

Homework Statement


You throw a baseball straight up in the air so it rises to a maximum height much greater than your height. Is the magnitude of acceleration greater while it is being thrown or after it leaves your hand. Explain.

Homework Equations


acceleration = velocity/time


The Attempt at a Solution


I know this is a really silly problem, but I just don't know the answer to it. The magnitude of the acceleration of the ball after it's left your hand would be due to gravity, which is 9.8 m/s^2. But is that more or less than the acceleration than the ball experiences while you throw it? Doesn't that completely depend on the amount of time it takes you from when your throwing arm is at rest to when it has a velocity and let's go of the ball?

But then they say that maximum height is "much" greater than my height. But doesn't that have to do with the velocity of the ball when it's left your hand, not the acceleration it experienced while you threw it?

I'm probably over-thinking this. Thanks
The formula for acceleration is delta V / delta T. What is the delta V and the delta T for each of the two legs of the ball's journey?
 
Think interms of forces. If the ball is simply sitting in your hand then the force of gravity is balanced with the force you hand is applying upward. Now if your hand is moving upward then what?
 
DaveC426913 said:
The formula for acceleration is delta V / delta T. What is the delta V and the delta T for each of the two legs of the ball's journey?

You mean numerically? Well, after you've thrown it it's -9.8 m/s^2, but the question phrases it in terms or magnitude, so isn't the magnitude always positive?
The delta V/delta T of the throwing seems unknowable to me with the information given to me in the problem, but it would be positive.
 
jedishrfu said:
Think interms of forces. If the ball is simply sitting in your hand then the force of gravity is balanced with the force you hand is applying upward. Now if your hand is moving upward then what?

Then yes there would be acceleration, but the question seems to me to be asking which magnitude is greater... right?
 
nhmllr said:
You mean numerically? Well, after you've thrown it it's -9.8 m/s^2, but the question phrases it in terms or magnitude, so isn't the magnitude always positive?
The delta V/delta T of the throwing seems unknowable to me with the information given to me in the problem, but it would be positive.
This is a conceptual question. If you're trying to apply numbers or magnitudes, you're missing the point of the question.

How long (time) did the ball accelerate while it was in the process of being thrown?

Not in terms of numbers, but as compared to how long was accelerating after it left your hand?It was thrown quite high, right? Therefore must have been quite a pitch.
 
DaveC426913 said:
This is a conceptual question. If you're trying to apply numbers or magnitudes, you're missing the point of the question.

How long (time) did the ball accelerate while it was in the process of being thrown?

Not in terms of numbers, but as compared to how long was accelerating after it left your hand?


It was thrown quite high, right? Therefore must have been quite a pitch.

The more I think about it... the more I think that it's when you're throwing it... right?
When I throw the ball, I accelerate it from 0 to some velocity in some amount of time. Then gravity accelerates the ball in the opposite direction until it's at zero again (which is the max height). The stretch of the ball's journey for which I am able to accelerate it is about equal to the length of my arm, which is less than my height. If the ball flies "much more than my height," that implies that it takes the accleration due to gravity a longer "stretch" (distance wise) to accelerate it in the opposite direction until it's at velocity = 0, than it took for ME, the thrower, to get it to the velocity that the ball traveled at when gravity took over.

In other words, me throwing it took a very little amount of distance and time, but the distance and time it took for the ball to "decelerate" up until the point it's at the max height is much greater.

Delta V is equal in both cases, but delta T should be much smaller for the portion where I throw it.

Okay... I think I got it. Thanks!
 
nhmllr said:
Okay... I think I got it. Thanks!
You're welcome. Nicely thought through, BTW.
 
Back
Top