A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically upward from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground.

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## 1 Answer

3+3x3-3+3=
3+3x3-3+3=

The equation for the first ball dropped under the force of gravity is s=gt^2/2 where g is the acceleration of gravity, s is the distance=h/2 and t the time taken to fall the distance s. For the second ball, s=vt-gt^2/2, where v is the vertical velocity which needs to overcome gravity. h/2=gt^2/2, so t=sqrt(h/g). Putting this value of t into the other equation we get h/2=vsqrt(h/g)-g*h/2g. vsqrt(h/g)=h and v=h/sqrt(h/g)=sqrt(h^2g/h)=sqrt(hg). The value of g is 32'/s^2 so  v=sqrt(hg)=4sqrt(2h), where h is in feet.

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