Asmall ball rolls horizontally off the edge of a tabletop of height h. it strikes the floor a distance x horizontally away from the edge of the table. (a) how long is the ball in the air? (b) what is its speed at the instant it leaves the table? give your answer in terms of the given variables and g.

All answers are assuming no air resistance.
a. You can use the equation s = ut + (1/2)at^2 to solve the first part.
as the initial vertical velocity is 0, ut = 0. Therefore, h = (1/2)gt^2.
Therefore t = (2gh)^(1/2).

b. Given that the ball does not slow down while in the air the ball will remain at a constant velocity, therefore x = ut, therefore u = x/((2gh)^(1/2))

There are you answers, I hope that helps

The answer is 4f======since electrostatic force is given by    we let f be the electrostatic force when  r = 1 m    if we let  denote the electrostatic force when the distance is decreased so that r = 0.5m, then    we established earlier that f = kq₁q₂. therefore    the force will now be 4f.