A dart player throws a dart horizontally at a velocity of 12.7 m/s. The dart hits the board 0.39 m below the height from which it was thrown.
How far away is the player from the board? Provide evidence.

Explanation:

Evidence? did someone measure something?

How about mathematical support.

The time needed to drop 0.39 m from vertical rest is

t =√(2h/g) = √(2(0.39)/9.8) = 0.28 s

d = vt = 12.7(0.28) = 3.58292... 3.6 m

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge

that would be because of both gravity and friction. so     c