Finding the speed of a metal ball

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To find the speed of a 0.500 kg metal ball just before it touches the floor after rolling off a table, the conservation of energy principle is applied. The initial kinetic energy (KEi) from the ball's speed of 5.00 m/s and the potential energy (PEi) from its height of 1.00 meter are combined. The equation KEf = PEi + KEi is used to solve for the final kinetic energy (KEf) just before impact. Since the ramp is frictionless, the ball accelerates down the incline without losing energy. The final speed can be calculated using KEf = 0.5mv^2.
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alright, so here's my question that i have to answer:
"A ball of mass .500kg is rolling across a table top with a speed of 5.00 m/s. When the ball reaches the edge of the table, it rolls down a frictionless incline onto the floor 1.00 meter below (without bouncing). What is the speed of the ball when it reaches the floor (just before it touches the floor)?"

I use the equation PEf + KEf = PEi + KEi and then get KEf = PEi, which then leads to
.5mv^2 = mgh
From here, I'm stuck. Since the ramp is frictionless, does the ball accelerate the same as it would if it was dropped? And, how would I add in the initial speed of 5.00 m/s?
 
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smithers11 said:
I use the equation PEf + KEf = PEi + KEi and then get KEf = PEi, which then leads to
.5mv^2 = mgh
But KEi ≠ 0.
From here, I'm stuck. Since the ramp is frictionless, does the ball accelerate the same as it would if it was dropped?
No, it slides down the incline.
And, how would I add in the initial speed of 5.00 m/s?
That speed will be reflected in KEi.
 
ok so then i would change my equation to KEf = PEi + KEi, and then from there use the equation KEf = .5mv^2 to solve for the velocity (speed), right?
 
Right.
 
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