# suppose that the three balls shown in figure below start simultaneously from the top of the hills. which one reaches the bottom first? explain.

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by anitha5

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by anitha5

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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.

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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.*

==========

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.

1) path AFB:

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) =

v2 = sqrt(2gH) = g sin Ф * t =>

3) path AEB:

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),

v1= √(2gH), t1 = √(2H/g), t2 = H/v1= √(H/2g)

t1+t2=

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.