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January 31, 2015

January 31, 2015

Posted by **DHilsath** on Monday, July 30, 2012 at 7:08am.

- Dynamics -
**Elena**, Monday, July 30, 2012 at 8:53amThe time of the first ball motion to the top point is

v=v(o1) –gt, v=0,

t=v(o1) /g=18/9.8 = 1.84 s.

The maximum height of the 1st ball is

H= v(o1) •t –gt²/2=16.53 m.

When the 2nd ball starts, the 1st ball is on its return trip during 2-1.84 =0.16 s.

It covers the distance hₒ=g(0.16)²/2 =0.125 m. and has the downward velocity

v(1)=g(0.16)=1.568 m/s.

At this instant the distance between two balls is 16.53-0.125=16.4 m, and two balls are moving towards each other: the first ball accelerates downward with initial velocity v(1)=1,568 m/s, the second ball decelerates upward at initial velocity v(2)=13.5 m/s. They cover the distances:

the1st ball h1= v1•t+gt²/2; the 2nd ball h2=v2•t-gt²/2.

h1+h2= 16.4 m.

=> v1•t+gt²/2+ v2•t-gt²/2 =16.4

t=16.4/(v1+v2)=16.4/(1.568+13.5)=1.088 s.

The point where the balls mett is located

h= v2•t-gt²/2 =13.5•1.088-g•(1.088)²/2 =8.89 m (above the ground)

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