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Draw acceleration time graph for an object thrown upward and reaches to a height h and comes back Only solutions should be posted |
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Answer» For the first half, when ball A GOES upV A =U A −gTV B =−gTV AB =V A −V B V AB =U.......(1)For the second half, when ball A comes downV A =−gTV B =0V AB =V A −V B V AB =−gT........(2)Hence from the equation (1) we observe that relative VELOCITY is independent of T. Which will be valid TILL ball A reaches maximum height point. It is given time taken in reaching h is t.Hence for the time t, V AB will be constant, whereas after time t, relative velocity changes its direction and increases afterwards. |
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