To solve the problem of finding the time taken for the ball to return to the platform (moving upwards with acceleration (a)), we can use the following reasoning:

Key Observations

• Let the ball be thrown upwards with velocity (u) relative to the platform.
• The platform accelerates upwards at (a), so in the platform’s frame (non-inertial), the ball experiences an effective downward acceleration of (g + a) (due to gravity (g) and pseudo force from the platform’s acceleration).

Equation of Motion

When the ball returns to the platform, its displacement relative to the platform is (0). Using the equation:
[ s = ut + \frac{1}{2}at^2 ]
Here:

• (s = 0) (displacement = 0),
• Initial velocity (u) (upwards),
• Acceleration (= -(g+a)) (downwards, hence negative).

Substitute values:
[ 0 = ut - \frac{1}{2}(g+a)t^2 ]

Factor out (t):
[ t\left(u - \frac{1}{2}(g+a)t\right) = 0 ]

Ignoring (t=0) (initial time), solve for (t):
[ u = \frac{1}{2}(g+a)t \implies t = \frac{2u}{g+a} ]

Answer: (\boxed{\dfrac{2u}{g+a}}) (assuming (u) is the velocity relative to the platform, (g) is gravity, (a) is platform’s upward acceleration).

If the problem uses specific symbols (e.g., (v) instead of (u)), replace (u) with the given variable. For example, if the ball’s velocity is (v), the answer is (\boxed{\dfrac{2v}{g+a}}).

Final Answer (using standard notation):
(\boxed{\dfrac{2u}{g+a}})

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