Problem:
A boy standing in a elevator accelerating upwards throws a ball upward with a velocity V. The ball returns to his hand after a time T. What is the acceleration of the lift?

Solution:
To solve this problem, we can use the equations of motion and the principles of relative motion.

Equations of motion:
1. v = u + at
2. s = ut + (1/2)at^2
3. v^2 = u^2 + 2as

Principles of relative motion:
1. If an object is thrown vertically upwards or downwards with a velocity V from a moving platform, its motion relative to the platform is the same as if it were thrown vertically upwards or downwards with a velocity V from a stationary platform.
2. If an object is thrown horizontally from a moving platform, its motion relative to the platform is the same as if it were thrown horizontally from a stationary platform with the same velocity.

Steps to solve the problem:
1. Let the initial velocity of the ball be u. Then, according to the first principle of relative motion, the velocity of the ball relative to the lift is u - V (since the lift is moving upwards with a velocity V).
2. When the ball returns to the boy's hand, its velocity relative to the lift is -u - V (since it is moving downwards with a velocity u and the lift is still moving upwards with a velocity V). The time taken for the ball to return to the boy's hand is T.
3. Using the first equation of motion, we can write:
-u - V = (u - V) + aT
=> a = -2u/T - 2V/T
4. We can use the third equation of motion to find the initial velocity of the ball. When the ball reaches its maximum height, its velocity becomes zero. Therefore, we have:
0 = u^2 - 2as
=> u^2 = 2as
5. Using the second equation of motion, we can write:
s = ut + (1/2)at^2
=> s = (u/2)(T) - (1/2)(2u/T + 2V/T)(T^2)
=> s = (u/2)T - uT - VT
6. Since the ball returns to the boy's hand, the displacement of the ball relative to the lift is zero. Therefore, we have:
s = 0
=> (u/2)T - uT - VT = 0
7. Using the equation u^2 = 2as, we can write:
u^2 = 2VT
=> u = sqrt(2VT)
8. Substituting the value of u in the equation (6), we get:
(sqrt(2VT)/2)T - 2VT - VT = 0
=> (sqrt(2VT)/2)T = 3VT
=> a = -2u/T - 2V/T = -3g, where g is the acceleration due to gravity.

Answer:
Therefore, the acceleration of the lift is -3g, where g is the acceleration due to gravity.

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