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

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When a ball is thrown upwards, it initially accelerates due to the force exerted by the hand, which is greater than the gravitational force. Upon release, the ball's acceleration changes to -g, meaning gravity is the only force acting on it. The ball continues to move upwards momentarily because it retains its initial upward velocity. There is no upward force once the ball leaves the hand; only gravity acts on it, causing it to eventually slow down and fall back down. This behavior aligns with Newton's first law of motion, which states that an object in motion will remain in motion unless acted upon by an external force.
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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