A particle is thrown upwards. It attains a height (h) after 5 seconds and again after 9s comes back. What is the speed of the particle at a height h?

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Abhinav Raj

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Total time taken by the particle to go up and come back = 5+9 =14s. Therefore it must take 7s to go up and 7s to come down. From first law, we know v=u+at At the top point velocity will be zero. Therefore v=0. 0=u-10*7 (taking a=g=-10m/s², negative sign indicates that acceleration is in downward...
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Total time taken by the particle to go up and come back = 5+9 =14s. Therefore it must take 7s to go up and 7s to come down. From first law, we know v=u+at At the top point velocity will be zero. Therefore v=0. 0=u-10*7 (taking a=g=-10m/s², negative sign indicates that acceleration is in downward direction). Solving the equation we will get u=70m/s. It is given that after 5s its height was h. Again applying first law. v=u+at v=70-10*5 v= 20m/s Particle speed was 20m/s in upward direction at height h after 5s of throwing. read less
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