An object of mass 2.00 kg is held at a position a, a vertical height of 20.0 m above the ground. point b is 8.00 m directly below a. neglect air resistance and use g = 10.0 m/s2. what is the speed of the object at position b when it falls from position a? i have no idea how to go about

find the time for the object to fall from position A to position B. You can use \[x = \frac{1}{2}at^2\]to do so. \(x\) represents the distance traveled, \(a\) is the acceleration (here due to gravity, so use the value provided for \(g\)), and \(t\) is the time. When you have found the time required for the object to fall from position A to position B, you can use \[v = at\] to calculate the speed. Again, \(a=g\) and \(t\) is the value you found in the first part ;)

Answer sample (c); // would be best because the percentage of the energy in an incident wave that remains in a reflected wave from this material is the smallest;

Newton's first law describes something special that results from the sum of all the external forces (including gravity) on an object.