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A body is released from top of tower of height h.It take t sec. to reach the ground.where will be the ball after time t/2 sec.

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one year ago

Saurabh Koranglekar
10233 Points
```							Dear studentS = ut + 0.5at^2U= 0 a= gt= ( 2s/g)^0.5t/ 2 = (s/2*g)^0.5S(@time = t/2) = 0.5*g*(s/2g)= s/4Or h/ 4Regards
```
one year ago
Arun
25211 Points
```							Let the acceleration be 'g'Then initial velocity = 0 m/s [in T seconds]Let the distance be 'h' m.We know,Equation of motion:s = ut + 1/2 gT²Then,h = 1/2gT² ......(1)h1 = 1/2gT²/4'......(2)Substitute (1) and (2):We'll get h1 = h4∴ The distance from ground is h-h/4= 3h/4
```
one year ago
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