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March 31, 2015

March 31, 2015

Posted by **Edward.Oki** on Sunday, March 24, 2013 at 2:58am.

- Physics -
**drwls**, Sunday, March 24, 2013 at 7:10amBall B will have to spend 1.5 seconds going up and 1.5 seconds coming down, since Ball A takes three seconds to go down. (Up and down times are equal)

Ball A's average velocity while going down is 6 m/s. Its final velocity is therefore 12 m/s, since it accelerates uniformly.

Let the distance Ball B travels up the ramp be x*18 m. (x is a dimensionless fraction)

Final velocity of Ball B:

VBmax = (18x/1.5)*2 = 24 x

Also,since the achieved maximum velocity is proportional to sqrtx,

VBmax = sqrt x*VAmax = sqrtx*12 m/s

24x = sqrtx*12

sqrtx = (1/2)

x = 1/4

Ball B travels 1/4 of the way up the ramp.

For the projection velocity of ball B,

VBmax = [sqrt(1/4)]*12 = 6 m/s.

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