Lift relative velocity question.

[takando12]

Homework Statement

From a lift moving upward with a uniform acceleration 'a', a man throws a ball vertically upward with a velocity v relative to the lift.The ball comes back to the man after a time 't' Show that a+g = 2v/t

Homework Equations

s=ut+1/2gt²

The Attempt at a Solution

I know this is a simple problem. By the substituting values 'v', 't' and a+g as acceleration in the above equation, i could get the answer. All of this is done by considering it from the lift's frame of reference. But I don't understand why we are adding g and a . SHouldn't we subtract them ? the ball goes up and down so must we subtract it the first time and add it the second? really confused.

 

[jbriggs444]

I know this is a simple problem. By the substituting values 'v', 't' and a+g as acceleration in the above equation, i could get the answer. All of this is done by considering it from the lift's frame of reference. But I don't understand why we are adding g and a . SHouldn't we subtract them ? the ball goes up and down so must we subtract it the first time and add it the second? really confused.

Yes, if one is computing the relative acceleration of two objects, one should take the difference of the two accelerations. But... take a look at the sign conventions here. The lift's acceleration is positive upward. The ball's acceleration (from gravity) is positive downward. Yes, you have to subtract. But before you subtract, you have give the two values the same sign convention. That means you have to negate one or the other. g-(-a) = g+a. That's the formal way to look at it.

The informal way to look at it is that the faster the elevator accelerates upward toward the ball, the faster the ball seems to accelerate downward toward the elevator. So the upward acceleration of the elevator adds to the downward acceleration of gravity.

 

[takando12]

thank you...i forgot the signs completely. And the informal explanation tops it nicely.