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?
[Physics] Direction of acceleration at highest point
accelerationforcesfree-body-diagramnewtonian-mechanics
Related Solutions
Can you have a zero velocity and nonzero average acceleration?
If you by velocity mean instantaneous velocity, then the question makes no sence. The corresponding acceleration will as well be instantaneous (and the answer would be yes.) $$\lim_{\Delta t \to 0}a_{av}=a_{inst}$$
- ... unless it is not a requirement that the acceleration is averaged over the same timespan as the velocity is measured. Then the answer is yes, and an upwards thrown stone is an example (it reaches a halt of $v=0$ and starts falling down, but the acceleration is at all times equal to the gravitational acceleration $-g$, so the average acceleration is as well, $a_{av}=-g$.)
If you mean average velocity, then the answer is no. Average acceleration doesn't take into account what happen in between; only the end points are interesting: $$a_{av}=\frac{v_2-v_1}{\Delta t}$$ If average acceleration is non-zero, then $v_1 \neq v_2$ and the average of these is surely non-zero as well.
You throw the ball upwards with velocity $v$ and it returns to your hand with velocity $-v$. Let's draw a graph showing the velocity as a function of time:
Acceleration is defined as:
$$ a = \frac{dv}{dt} $$
so it is the gradient of the line in this graph. The velocity-time line is straight so the gradient is constant which means the acceleration is constant. The gradient is just the gravitational acceleration $9.81$ m/s$^2$.
The point is that the gradient, and hence the acceleration, does not depend on $v$ at all. So it is the same value of $9.81$ m/s$^2$ when $v = 0$ just as it is at all other values of $v$.
Best Answer
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.