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2026年质量好的pp永磁离心风机厂家选购指南与推荐-熙诚环保科技(苏州)有限公司

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}})

熙诚环保科技(苏州)有限公司

熙诚环保科技(苏州)有限公司



作者声明:本文包含人工智能生成内容。

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