Is the frame of elevator Preferred over the frame of ground? When an object is thrown upward with some velocity then it doesn't fall towards the earth until the given velocity becomes zero.
Now imagine the case given in the figure 
There is a question in my book asking for the time taken by the block to slide down the incline if it is released from the top of the incline. And the given length of the incline is $l$ .
There are two ways to solve this. One from the frame of elevator and the other from the frame of earth.
The first is to solve the question from the frame of elevator and in this process the body can be thought of to be released from rest and the time can be calculated using
$l = \frac{ 1}{2} gt^2$ ........(1)
But my doubt is with second process. As seen from the frame of earth, we will see that the body has a component of velocity $v\sin (\theta)$along the incline but in the opposite direction of the component of acceleration due to gravity ($g\sin(\theta)$) along the incline i.e. in upward direction.
Now applying the concept written in the beginning lines of this question the body shouldn't fall along the incline unless and until the component of velocity given in the opposite direction becomes zero. And since the elevator doesn't stop the body shouldn't touch the bottom of the incline ever.
Edit: The time taken by body as seen from ground frame will be :
$l = (u\sin \theta)t - \frac{1}{2} gt^2$........(2)
In both the equations (1) and (2) time taken is different.
But how is this possible?? Aren't both the frames equivalent ?
Can anyone give a reasonable answer ?
Please help me with this.
 A: 
Now applying the concept written in the first para of this question the body shouldn't fall along the incline unless and until the component of velocity given in the opposite direction becomes zero

This concept does not apply to this scenario. In fact, the failure of that line of reasoning is exactly the indication that it does not apply.
Many statements that you can make about physical systems are coordinate dependent statements. Such statements generally will only work in certain coordinate systems. This is one of them.
In this case the issue is that your statement is irrelevant in the second scenario:

When an object is thrown upward with some velocity then it doesn't fall towards the earth until the given velocity becomes zero.

For this problem the motion of the block with respect to the earth is irrelevant. The block may be sliding to the bottom of the ramp at any speed relative to the earth for some range of $v$. So it is of no interest to the problem when or if the block comes to rest in the earth’s frame.
In general, you are free to use any inertial frame to solve a problem. Therefore, you should always strive to use the frame that makes it easiest. In this case, the easiest frame is the frame of the elevator.
A: In the ground frame, the block will be moving upwards overall even though it falls relative to the incline. The net force directed along the incline will slow down the upward velocity of the block (while also making it gain a horizontal velocity).
So, in short, the block does not move downwards when seen from the ground frame. Instead, its upward velocity keeps decreasing.
So why does the block fall along the incline even in the ground frame? Because, in the ground frame, the block is moving with a less upward velocity than the incline is. Initially they were both moving with the same upward velocity, but there is a decelerating force on the block, while there is no such force on the incline.
The block does not fall overall in the ground frame (instead, it moves upwards). It only falls relative to the incline (as they have a difference in upward velocities)
EDIT- In the elevator frame:
I'll assume you've set up your x-axis as along the incline, and y-axis as perpendicular to the incline. And the block is initially at the origin.
The initial velocity is 0. The acceleration perpendicular to the incline is 0 (as there is no motion perpendicular to the incline). The acceleration along the incline is $-g\sin{\theta}$. The displacement is $-l$ (we can take whichever direction to be negative). The time taken is given by:
$-l=-\frac{1}{2}g\sin{\theta}t^2$
In the ground frame:
Wel'll have to attach our co-ordinate axes to the ground. Again, our x-axis is parallel to the incline, and y axis is perpendicular to the incline. But this time, the co-ordinate system is fixed to the ground.
When the block will have reached the bottom of the incline, it'll have a total displacement of of $u\sin{\theta}t-l$ along the x-axis (axis parallel to the incline), when seen from the ground frame. The acceleration along the incline is again $-g\sin{\theta}$. So we get:
$u\sin{\theta}t-l=u\sin{\theta}t-\frac{1}{2}g\sin{\theta}t^2$
This gives the same value of $t$ as in the elevator frame.
Math aside, key point is that the block obviously does not travel a distance of $l$ when seen from the ground frame. In the ground frame, block also has some initial velocity $u\sin{\theta}$ along the incline. This leads to a net displacement of $u\sin{\theta}t-l$ in the direction of the incline.
A: Both frames are inertial, so you can apply Newtons second law in a similar fashion in each frame.  Since the two frames have no relative acceleration, the components of acceleration for the block must be the same in both frames.  Since both ends of the slope are rising at the same rate, the effective angle and length of the slope does not change. You get the same slide time in either frame.
