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It seems that the ball in the middle one takes the least time and reaches the bottom first.  But the velocity at the bottom is the same.

See enclosed diagram with this answer.

     For simplicity, instead of curved paths, let us take paths with two inclines.   Let us calculate the time of travel for the ball from A to B on various paths in the diagram. If we use a very large number of inclined segments, then we get the curved path. The result obtained using multiple linear segments is applicable for the curved paths.

1) path AFB:
   angles: tanФ1=1/3, tanФ1=3, sinФ1= 1/√10, sinФ1=3/√10

v1 at F   = g sinФ1 * t1,   v1^2 = 2g H/4 = gH/2,       v1 = √(gH/2)
t1 = √(H/2g)/ (Sin Ф1)               t1 = √10 * sqrt(H/2g)
v2^2 = v1^2+2g 3H/4 = 2gH       sqrt(2gH) = sqrt(gH/2) + g sin Ф2 * t2
t2 = (√10/3) √(H/2g) = (√10/3) √(H/2g)
t1+ t2 = 2/3 * sqrt(10)* sqrt(2H/g) = 2.108 * √(2H/g)

2) path ADB:
    angle  tan Ф = 1       sin Ф = 1/√2
v2 = sqrt(2gH) = g sin Ф * t         => t = 1.414  (2H/g) 

3) path AEB:
  angle: tanФ1=3, tanФ2= 1/3, sinФ1= 3/√10, tanФ2=1/√10
v1 = g sinФ1 *t1 = √(2g3H/4),             t1 = √10 * √(H/2g) /√3
v2 = √(2gH) = √(3gH/2) + g sinФ2* t2,        t2 = √10 √(H/2g) [2-√3]
t1+t2 = √10 * √(2H/g) (1 - 1/√3),    t = 1.33  √(2H/g)

4)  path AOB:  First vertical and then horizontal.
 v1= √(2gH),       t1 = √(2H/g),         t2 = H/v1= √(H/2g)
t1+t2= 1.5 * √(2H/g) 

    So we can see that the ball along the path AFB takes longest time to reach B.  The ball along single straight line path, ADB takes less time duration to reach B.  For the path AEB, the time duration reduces further to a minimum.  But after a particular point, as the slope of AE increases, the slope of EB reduces and the total time duration to reach B increases.

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