A ball is thrown and I'm trying to understand why

[exparrot]

Hi everyone, I'm new to this forum. Just would like some help since I'm trying to wrap my brain around this scenario:

A ball is thrown upward. While the ball is in the air, does its acceleration increase, decrease, or remain the same? Describe what happens to the velocity of the object from when it is thrown until it returns.

So I'm thinking that since this is a free fall scenario, the ball will reach a maximum height (at that point the velocity is 0 m/s), after which acceleration due to gravity will act on the ball. I'm not sure why any of the three above possibilities could happen. I mean, it seems like velocity would increase, thus acceleration would increase. I'm not sure if that is the correct assumption. I would appreciate if anyone could help me out with understanding this question. Thanks!

 

[rock.freak667]

So I'm thinking that since this is a free fall scenario, the ball will reach a maximum height (at that point the velocity is 0 m/s), after which acceleration due to gravity will act on the ball. I'm not sure why any of the three above possibilities could happen. I mean, it seems like velocity would increase, thus acceleration would increase. I'm not sure if that is the correct assumption. I would appreciate if anyone could help me out with understanding this question. Thanks!

For free fall motion, ignoring friction, there is only one acceleration and that is gravity. On the way up, the acceleration is -g and on the way down it's +g.

EDIT: listen to russ, I sort of changed convention for two different parts of motion.

 

[russ_watters]

That's a confusing way to say it rock freak. It is best to consider it a continuous motion with constant acceleration of -g rather than a positive speed up and a positive speed down.

 

[DaveC426913]

Yep. Constant negative acceleration. Velocity starts off positive, ends up negative.

The acceleration graph would be a straight line below the x-axis.

 |
 |
0|...
 |
 |_________

The velocity graph would be a straight negative slope.

 |\
 | \
0|..\...
 |   \
 |    \

 

[RohansK]

The velocity of the ball would continously reduce as it travels upwards, becomes Zero at the top most position and the ball staarts the downward fall. As the ball travels upward its positive acceleeration goes on reducing under the influence of gravity.

And as the ball starts the fre fall under gravity, the gravirational acceleration remains cosntant and acting downwards ( negative) so the velocity continously increases. This is because constant acceleration means a CONSTANT RATE OF CHANGE in velocity . Thus the velocity keps on inceasing as the ball continues is downward fall.

 

[exparrot]

Thank you all for answering. I've got a better understanding of this concept now!

 

[DaveC426913]

The velocity of the ball would continously reduce as it travels upwards, becomes Zero at the top most position and the ball staarts the downward fall. As the ball travels upward its positive acceleeration goes on reducing under the influence of gravity.

And as the ball starts the fre fall under gravity, the gravirational acceleration remains cosntant and acting downwards ( negative) so the velocity continously increases. This is because constant acceleration means a CONSTANT RATE OF CHANGE in velocity . Thus the velocity keps on inceasing as the ball continues is downward fall.

You have made the same mistake as rock freak. The acceleration does not change from positive to negative; it is constant from toss right through to landing. (By convention up is positive, thus the acceleration is constant and negative.)

 

[mikelepore]

The vertical component of acceleration is -g (negative) during both the upward and downward motions. You have negative acceleration when you're going in the positive direction and slowing down, and when you're when going in the negative direction and speeding up.

 

[RohansK]

Right Dave, thanks for correcting me, I had the conventionals signs in my mind.

But let me take the opportunity to ask a small doubt I always had. Taking the same example above, when we say that the ball is thrown upwards the ball does gain speed from zero to the maximum velocity at beginning ( velocity with which it leaves the hand). So can we say tht the initial velocity ( at the instant ball leaves the hand) is suppose 20m/s, then can we say that the initial upward acceleration is ( 20-0)/t where t is the time from which the hand starts from rest holding the ball to the time the ball is released from the hand. would it be correct.
Or we only consider the accelration due to gravity (-g) continously acting on the ball rfom the time it leaves the hand, where the velocity is maximum. How do we describe the acceleration of the ball from zero to V max when it leaves the hand.

can you please clear this doubt for me.

 

[berkeman]

Right Dave, thanks for correcting me, I had the conventionals signs in my mind.

But let me take the opportunity to ask a small doubt I always had. Taking the same example above, when we say that the ball is thrown upwards the ball does gain speed from zero to the maximum velocity at beginning ( velocity with which it leaves the hand). So can we say tht the initial velocity ( at the instant ball leaves the hand) is suppose 20m/s, then can we say that the initial upward acceleration is ( 20-0)/t where t is the time from which the hand starts from rest holding the ball to the time the ball is released from the hand. would it be correct.
Or we only consider the accelration due to gravity (-g) continously acting on the ball rfom the time it leaves the hand, where the velocity is maximum. How do we describe the acceleration of the ball from zero to V max when it leaves the hand.

can you please clear this doubt for me.

In simplified questions like this one, the acceleration to the release velocity is not part of the equations. Of course in the real world it would be, but if defined well, it would not affect the final answer.

But in a rocketry question, then you would need to take lots of stuff into account, eh?