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## Homework Statement

An elevator is going up from the ground with acceleration a

_{elevator/ground}.

When the elevator's height measured from the ground is h, its velocity us v

_{elevator/ground}(assume that t = 0 in this condition), a ball is thrown up with velocity v

_{ball/elevator}relative to the elevator. The gravitational acceleration is g.

What is the acceleration of the ball with respect to the ground ?

In what time does the ball height get maximum measured from the ground ?

**2. Relevant equation**

v(t) = v(0) + a t

v(t) = v(0) + a t

## The Attempt at a Solution

I thought that the ball acceleration relative to the ground is a

_{elevator/ground/}- g

And, the ball velocity relative to the ground is V

_{b/e}+V

_{e/g}

So, the time it takes to get maximum height is :

0 = V

_{b/e}+V

_{e/g}+ (a

_{elevator/ground}- g) t

t = - (V

_{b/e}+V

_{e/g}) / (a

_{elevator/ground}- g)

t = (V

_{b/e}+V

_{e/g}) / (g - a

_{elevator/ground})

Am I right ?

I'm in doubt because someone in yahoo answer says that the acceleration is just -g (see https://id.answers.yahoo.com/question/index?qid=20101112142831AAzmKKU ). It is in Bahasa Indonesia, not English)