Why does a ball continue to travel upwards when thrown into the air?

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SUMMARY

The discussion clarifies the mechanics of a ball thrown upwards, emphasizing that once released, the only force acting on the ball is gravity, which accelerates it downwards at -9.81 m/s². Initially, the ball accelerates upwards due to the force exerted by the hand, but upon release, it retains its upward velocity until gravity overcomes it. The key takeaway is that there is no upward force acting on the ball after it leaves the hand, and it will continue to ascend only until gravity decelerates it to a stop before pulling it back down.

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tyneoh
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Hello everyone, I am have some confusion regarding gravity and a ball traveling upwards.

Suppose you throw a ball upwards into the air. At the beginning, the ball is at rest atop your hand, Freaction=mg. When your hand moves upwards to throw the ball, Freaction>mg and the ball accelerates upwards with acceleration a,while in contact with your hand. Upon release, the ball will travel upwards with the same acceleration.

But with the increase in height, a will decrease. I know that this is due to the pull of gravity but I can't seem to understand fully as when the ball soars, the upward thrust is greater than that of the weight of the ball, mg.

Thus if the thrust is greater than the pull of gravity on the ball, there is a net force acting upwards, propelling the ball skywards perpetually, which is illogical.

Could somebody please correct my reasoning and end my confusion? Many thanks in advance :)
 
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tyneoh said:
When your hand moves upwards to throw the ball, Freaction>mg and the ball accelerates upwards with acceleration a,while in contact with your hand. Upon release, the ball will travel upwards with the same acceleration.
No. Upon release, the acceleration of the ball drops to -g.

tyneoh said:
But with the increase in height, a will decrease.
No, the velocity v will decrease. The acceleration during flight is constant -g (ignoring drag)

tyneoh said:
Thus if the thrust is greater than the pull of gravity on the ball,
There is no thrust once the ball leaves the hand. Just gravity.
 
A.T. said:
No. Upon release, the acceleration of the ball drops to -g.


No, the velocity v will decrease. The acceleration during flight is constant -g (ignoring drag)


There is no thrust once the ball leaves the hand. Just gravity.

I am starting to see your point here, but allow me to ask: when the ball leaves the hand, it will still travel upwards, right? Shouldn't the be a upward force compelling it to move?
 
tyneoh said:
I am starting to see your point here, but allow me to ask: when the ball leaves the hand, it will still travel upwards, right? Shouldn't the be a upward force compelling it to move?

No. In the absence of force the ball will continue to move at the same speed, in the same direction, forever. This principal is called momentum. Once the ball leaves your hand the only force acting on it is gravity, which is why it comes down. If not for gravity it would continue upward into the universe forever

All this assumes no air resistance.
 
tyneoh said:
I am starting to see your point here, but allow me to ask: when the ball leaves the hand, it will still travel upwards, right?
Sure, since it has an initial upward velocity.
Shouldn't the be a upward force compelling it to move?
No. In the absence of force, things keep moving at the same velocity. (Newton's 1st law.) Force is needed to change velocity, not maintain it. Once it leaves your hand, the only force acting on the ball is gravity, which accelerates the ball downward.
 

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