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Asked by Sudha Last Modified
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 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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