let the total distance be s so, s/2 is covered in 1 second so, initial velocity at the start of that 1 sec. be u =>s/2 = u + 1/2g => u = (s+g)/2 let us consider the body was just dropped, so initial velocity at start of motion = 0
=> u = gt where t is time taken to travel 1st half distance => t = u/g = (s+g)/2g ----(1)
now, s/2 = 1/2gt^2 => s = gt^2 putting this in equation (1) & solving we get t = 1 sec.
Moderation Team
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