Acceleration of a ball thrown vertically upwards

[Janiceleong26]

Homework Statement

I chose A because since the acceleration of the ball is positive vertically upwards, the acceleration downwards is negative, and at the turning point, acceleration is zero.. The correct answer is B. Why?

 

[Doc Al]

I chose A because since the acceleration of the ball is positive vertically upwards, the acceleration downwards is negative, and at the turning point, acceleration is zero..

Perhaps you are confusing acceleration (which is the rate of change in velocity) with velocity? Why do you think the acceleration changes? Does the force change?

 

[Doc Al]

Think about this:

If the acceleration were 0 at the maximum height, the ball would just sit there suspended. Does that happen?

If the acceleration were positive as the ball rises, then it would get faster and faster as it does. Does that happen?

 

[Janiceleong26]

Think about this:

If the acceleration were 0 at the maximum height, the ball would just sit there suspended. Does that happen?

If the acceleration were positive as the ball rises, then it would get faster and faster as it does. Does that happen?

Oh..I get what you mean.. I thought acceleration was zero at the maximum height, because v=0.. but I shouldn't see into velocity, it's the change, thanks for the help!

 

[Janiceleong26]

Perhaps you are confusing acceleration (which is the rate of change in velocity) with velocity? Why do you think the acceleration changes? Does the force change?

But why, whenever we do Kinematics questions, where an object falls, we take acceleration as g= + 9.81ms^-2 ?

 

[Doc Al]

But why, whenever we do Kinematics questions, where an object falls, we take acceleration as g= + 9.81ms^-2 ?

I'm not exactly sure what you are asking, so I'll mention two things:
(1) g is usually taken as a positive constant; g = 9.81ms^-2;
(2) for a falling body, the acceleration is downward with magnitude g; whether you express that downward acceleration as + or - depends on your sign convention. (Sometimes it's convenient to use "down = positive".)

 

[Janiceleong26]

I'm not exactly sure what you are asking, so I'll mention two things:
(1) g is usually taken as a positive constant; g = 9.81ms^-2;
(2) for a falling body, the acceleration is downward with magnitude g; whether you express that downward acceleration as + or - depends on your sign convention. (Sometimes it's convenient to use "down = positive".)

Ok thanks !