Ball in Free-Fall: Find Final Speed

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The discussion focuses on calculating the final speed of a ball dropped from a height of 13.9 m, with an initial downward speed of 4.9 m/s, when it hits a table 0.7 m below. For part (a), the calculated final speed is 16.8 m/s, while for part (b), where the ball is thrown upwards at the same speed, the final speed is 17.3 m/s. The equation used for these calculations is vf = sqrt(v(initial)^2 + 2g(h - y), with the distance traveled being 0.7 m. There is some confusion regarding the term "dropped from rest" given the initial velocity provided. The key takeaway is that the initial velocity must be considered in both scenarios to determine the correct final speeds.
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Homework Statement



A ball of mass m = 8 kg is dropped from rest at a height h = 13.9 m above the ground. Ignore air resistance.

(a)If the ball is being released with a downward speed 4.9 m/s initially, what will be its final speed when it hits the table 0.7 m below the release point?

(b) If the ball is being thrown upwards instead with the same speed 4.9 m/s, what is the final speed when it hits the table?

Homework Equations



Kf + Ugf = Ki + Ugi

The Attempt at a Solution


vf = sqrt(v(initial)^2 + 2g(h - y))
I just plugged in numbers and got 16.8 for part (a) and I plugged in numbers for part (b) and got 17.3. Also, for part (b), I substituted y=0 because I chose the y-axis to start at the bottom where the hand would be.
 
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To be clear. The ball is released at 13.9m and it hits the table at .7m?

And the initial velocity is 4.9m/s.

What does dropped from rest mean when you have an initial velocity given?
 
vf = sqrt(v(initial)^2 + 2g(h - y))
In the first part the distance traveled by the ball is only 0.7m.
 
Ignore that released from rest part.
 
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