A ball is dropped from pole ABC where B is its midpoint so AB=BC.find the ratio of time of descent from AB to BC.
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we apply the equation : s = u t + 1/2 a t² , here u = 0 and a = g
AB = 1/2 g t1² => t1 = √(2 AB/ g)
and v = g t1 = √(2 g AB)
For descent from AC :
AC = 2 AB = 1/2 g t2² => t2 = 2 √(AB /g)
Time of descent from B to C = t2 - t1 = (2 - √2) √(AB/g)
Ratio of time durations = t1 : t2-t1
= √2 : (2 - √2)
= 1 : (√2 -1)
AB = 1/2 g t1² => t1 = √(2 AB/ g)
and v = g t1 = √(2 g AB)
For descent from AC :
AC = 2 AB = 1/2 g t2² => t2 = 2 √(AB /g)
Time of descent from B to C = t2 - t1 = (2 - √2) √(AB/g)
Ratio of time durations = t1 : t2-t1
= √2 : (2 - √2)
= 1 : (√2 -1)
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