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Grade: 12th pass
        
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.
3 months ago

Answers : (2)

Saurabh Koranglekar
askIITians Faculty
3138 Points
							Dear student

S = ut + 0.5at^2
U= 0 a= g

t= ( 2s/g)^0.5

t/ 2 = (s/2*g)^0.5

S(@time = t/2) = 0.5*g*(s/2g)= s/4

Or h/ 4

Regards
3 months ago
Arun
22533 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 gT²
Then,
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
3 months ago
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