Another simple physics question

do you mean straight upwards away from the earth's surface?

 

If you mean throwing it straight up from the earth's surface, it will technically:

A)start decelerating as soon as the force is removed (as soon as it leaves your hand if you are throwing it),

B)stop as it reaches it's pinnacle (gravity overcomes it's upward movement) and

C) accelerate at the rate of gravity as it falls back to earth.'

 

i would say (A) due to the gravity just affecting the amount acceleration, althuogh the acceleration is constant, dont know if i said that right, but it shard to put into words at the moment

 

Yeah what he said. Assuming it is being moved initially away from the surface of earth.

 

Gotcha...... After it leaves your hand both you and and the ball are undergoing 1G of acceleration from gravity. Did I miss something?

 

you may have just gave it away there..

 

C. Decelerate from the point thrown till it reaches it's apex thereby losing it's fight with gravity after which it will accelerate back towards the earth.

 

god I hope not, there is nothing more to discuss, if people don't understand basic principles of gravity and accel/decel, well they shouldn't even bother posting.

 

The only confusing piece is "at the earths surface" As I said my reply was assuming it was being thrown away from the earths surface. If you are throwing it towards the ground then it would be quite the opposite.

I'm failing to see how you can throw it UP AT the earths surface?

All this funky ass wording, my sumation was correct, but you may need to clarify the question.

 

The direction of the acceleration vector does not change with respect to the person who threw the object as long as that person's postion is fixed relative to the center of mass of the earth. The object will continue to accelerate toward the center of mass of the earth at apx 9.8m/s2 until it hits the ground. If we define deceleration as negative acceleration with respect to velocity, then a object thrown straight up at 10m/s will decelerate for slightly more than 1 s. When its velocity reaches zero, it will begin accelerating until it hits the ground.
If you are standing on the surface of the earth, you are not experiencing any gavitational acceleration; the force of gravity is exactly countered by the normal force exerted on your feet by the ground on which you are standing.
You are experiencing a centripedal acceleration due to the Earth's rotation (unless you are standing on one of the rotational, as opposed to magnetic, poles). If you through the ball such that it was farther from the equitorial plane than you were, then there would be a net corriolis force on the ball with respect to you, but this would be exceedingly small. It is simply a myth that corriolis force causes small drains (eg sinks) to produce counter clockwise vortices in the nothern hemisphere and clockwise vortices in the southern hemisphere; it is just too small. You need a radius of many meters before the corriolis force on earth has a noticable effect on water vortices.

 

I voted C


Antony

 

Sorry, I voted for accellerate, then decelerate before reading it is being thrown away from the earth's surface. Upon reading the question I assumed a person was essentially in the air upside down (thus being able to throw up at the surface of the earth) throwing towards the earth's surface.
BT

 

So now understanding the question it would decelerate, then stop, then accelerate, then stop.
BT