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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? Follow 1  Answer 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...

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.

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