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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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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


At the top point velocity will be zero. Therefore v=0.


(taking a=g=-10m/s²,  negative sign indicates that acceleration is in downward direction). Solving the equation we will get


It is given that after 5s its height was h. 

Again applying first law. 



v= 20m/s

Particle speed was 20m/s in upward direction at height h after 5s of throwing. 

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