An elevator is going up from the ground with acceleration aelevator/ground.
When the elevator's height measured from the ground is h, its velocity us velevator/ground (assume that t = 0 in this condition), a ball is thrown up with velocity vball/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
The Attempt at a Solution
I thought that the ball acceleration relative to the ground is aelevator/ground/ - g
And, the ball velocity relative to the ground is Vb/e+Ve/g
So, the time it takes to get maximum height is :
0 = Vb/e+Ve/g + (aelevator/ground - g) t
t = - (Vb/e+Ve/g) / (aelevator/ground - g)
t = (Vb/e+Ve/g) / (g - aelevator/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)