When a ball is thrown vertically upwards, what will be the direction of acceleration at the highest point(where velocity is zero)? Upwards, downwards or arbitrary?
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$\begingroup$ The ball is only "at the highest point" for a very brief time. Before that is is still going toward the top or already falling back. So "at the highest point" would be at a lenggth of 0 seconds in time. Is anything experiencing anything in 0 seconds? $\endgroup$– HennesCommented Jun 7, 2015 at 13:50
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$\begingroup$ @Hennes It could experience it in micro seconds(maybe). $\endgroup$– MustagheesCommented Jun 7, 2015 at 13:55
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$\begingroup$ Yes, but that is not at the top. For this question: I would say: 1) A is always down (always to the center of the earth) 2) It is 0, but is "+0" the same as "-0"? (in my head it is the same, but this may be philosofical). 3) From a math perspective, it may be positive is reaching the limit from one side and negetive is reaching that point from the other side. (But in this specific case, positive-0-positive). $\endgroup$– HennesCommented Jun 7, 2015 at 13:59
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$\begingroup$ @Hennes if the ball experiences no acceleration at the topmost point, what causes it to fall? $\endgroup$– AsherCommented Jun 7, 2015 at 17:20
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$\begingroup$ In no time nothing falls. $\endgroup$– HennesCommented Jun 7, 2015 at 17:40
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4 Answers
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During the toss, while still in your hand, the ball is accelerated upwards.
As soon as the ball leaves your hand, it begins to slow down. Here, scientific and common usage diverge. Technically, the ball experiences a downward force, and its velocity decreases with time, so it can be said to possess a negative (downwards) acceleration. In common usage, though, when acceleration reduces speed (magnitude of velocity), this is referred to as deceleration. Once the ball reaches its peak, it will begin to fall downwards, and at this point both technical and common use agree - it is accelerating downwards.
So the question becomes, what exactly did you mean by your question? Technically, the ball experiences a constant negative (downwards) acceleration. In common use, the ball decelerates until its peak, then accelerates downwards. In this sense, at peak altitude the ball accelerates downwards. If it did not, it would remain at that altitude forever, since its vertical velocity would be zero, and it would not experience any change in that velocity. The fact that, just for an instant, the ball has zero velocity, does not mean that the ball is not accelerating - it just means that the velocity is zero.
answered Jun 7, 2015 at 13:33
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The accelaration is only because of the gravity, that is downwards. So when the ball reaches at the top of it's trajectory and it's speed becomes zero, it also has an accelaration pointing down.
After that the ball accelerates until it reaches ground(if there wasn't a ground it would oscillate around the point 0 from were the gravity accelaration is asked.)
Hope this helps.
answered Jun 7, 2015 at 13:16
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$\begingroup$ So at the top, is acceleration 0 or downwards? $\endgroup$ Commented Jun 7, 2015 at 13:19
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1$\begingroup$ The acceleration is downwards from the moment it leaves your hand to the moment it touches the ground. $\endgroup$ Commented Jun 7, 2015 at 13:20
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$\begingroup$ Made an edit, the comment was correct. I though of a force being applied to the ball for a distance, with the force getting smaller. So, the accelaration is downwards. $\endgroup$ Commented Jun 7, 2015 at 13:25
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By applying the equation $v=u+at$ ($v$ is final velocity, $u$ is initial velocity, $a$ is acceleration and $t$ is time) here and taking $a_g = -9.8 \,\mathrm{m/s^2}$ (at this point acceleration due to gravity is in opposite direction of motion) we see that the value of $v$ slowly decreases with time. At a certain point it becomes $0$. The ball is at $0$ velocity at that instant (instant is very important). After that the velocity is not strong enough to go against gravity and move up and so loses infront of gravity and begins to obey its words and come down along with it.
Remember that direction of acceleration due to gravity is always $9.8\,\mathrm{m/s^2}$ towards the center of the earth as stated in Newton's Shell Theorem (Please read it to get a better understanding).
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The question is ill-formed. Even if you throw the ball along a line extending thru the center of the earth ("perfectly vertical"), there'll be a transverse velocity due to you and the earth rotating. The ball will have a transverse velocity at its apex which will appear to you the observer as tho the ball is travelling to the west. This is because it's at a greater radius than when on the ground, so its transverse speed is insufficient to travel the same angular distance as you do.
Granted, you probably meant this to be a "idealized" situation w/ a uniform, flat world with no motion, but keep in mind that a real-world experiment will work out differently.
answered Jun 7, 2015 at 17:46
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$\begingroup$ Yes I had a simple situation in mind, you made it complex. $\endgroup$ Commented Jun 8, 2015 at 0:59