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With what speed will the ball hit the floor?

A child's toy consists of a block that attaches to a table with a suction cup, a spring connected to that block, a ball, and a launching ramp. The spring has a spring constant "k" , the ball has a mass "m", and the ramp rises a height "y" above the table, the surface of which is a height "H" above the floor.

Initially, the spring rests at its equilibrium length. The spring then is compressed a distance "s" , where the ball is held at rest. The ball is then released, launching it up the ramp. When the ball leaves the launching ramp its velocity vector makes an angle "delta" with respect to the horizontal.

Throughout this problem, ignore friction and air resistance.

express the speed in terms of "k," "s," "m," "g," "y," and/or "H"


...that would be the speed when the ball leaves the launching ramp. i was looking for the speed when the ball hits the floor..thanks

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The speed at the bottom of the table is sq((ks^2+2mgH)/m). To get the speed at the bottom of the table i used the formula (1/2)ks^2+mgH=(1/2)mv^2 then solve for v (disregard positive or negative signs since you only want speed). (1/2)ks^2 is the potential energy of the spring and mgH is the potential energy from gravity, which is weight of the ball (mg) times the height off the ground.

note: the height of the ramp is not included in the calculations since when dealing with energy all the information that is needed is the beginning state and the ending state, conservation of energy.

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