Which is greater?

Question

You throw a ball straight up in the air so that it reaches a maximum height of 14.5 m. Is the magnitude of its acceleration greater while it is being thrown or after it leaves your hand?

Answer 1

The answers so far have been atrocious. They will confuse or (hopefully) just amuse you.

If it reaches a height of 14.5m, it has a potential energy of mgh=14.5m* 9.8m/s^2*m. That means it had the same kinetic energy when you released it: 1/2mv^2=142.1*m J/kg implies v= 16.9 m/s. Now, after it has been released the acceleration is a constant 9.8m/s^2 downward (the only force on it is gravity). Now, to find the acceleration while it's in your hand, you have to make some assumptions, so technically there is not enough information to do this problem. However, with the reasonable assumption that your arm is about a meter long, if you accelerate the ball upward at 9.8m/s, it won't reach the necessary velocity by the time you release it. v=a*t and x=1/2*a*t^2 yield v=2*sqrt(x/a)*a=6.26m/s. So even in the ideal case, unless you are over 20 feet tall, the acceleration must be larger (at some point) while it's in your hand than while it's in the air. Your teacher has either given an intentionally tricky question, has made a mistake. If the later is the case, I'd be happy to personally call him/her to straighten things out.

Also, don't forget that the magnitude of the acceleration doesn't have a direction, so that part isn't important.

--- Update----
Sorry, I forgot a factor of 1/2 the first time......

Answer 2

The acceleration is almost invariably greatest at some point during the process of throwing the ball.

Reason being is that as your arm and hand are putting momentum into the ball, there will be a peak in that momentum. And unless the ball is released PRECISELY at that moment of maximum speed, then you have released the ball at some lesser speed.

Releasing the ball precisely at that moment is unlikely and a most improbable probability.

Also, once the ball has left your hand there will be no more acceleration unless some other force were to continue increasing it, such as if you replaced the ball with a bottle rocket, where the "rocket" continued to give it even more energy.

Answer 3

The greatest accelaration is probably negative and at greatest just before it reaches the ground. If you throw it up then there is the acceleration you have put on the ball minus the accelartion due to gravity whereas on its downfall there is only the earths gravitational effect on the ball.

Between the two suggestions you have offered it is definitely under the greatest accelertion during the throw. The moment you let go it is decelerating due to the effect of gravity on earth.

Answer 4

The instantaneous acceleration as it is released from your hand is the greatest acceleration that will be attained by the ball on its total journey up and down.

Remember in order for it to go up you have thrown it upward with an acceleration greater than gravity and as it goes up both air friction and gravity work in conjunction to counteract its ascent.

Answer 5

The acceleration on the ball just prior to release is the vector sum of gravity and centripetal acceleration. After the release, only gravity is operating, so the magnitude of the acceleration on the ball is greatest just before release.

Answer 6

once it leaves your hand it is no longer positively accelerating...so i would think while its being thrown

Answer 7

It will begin decelerating as soon as you let go of it, because of gravity and air resistance.

Answer 8

after it leaves your hand I think but could use more details?

Answer 9

Why don't you calculate it, graph it, and see?

Answer 10

neither, it's the same