Let the ball is thrown horizontally with some speed "v" and its height from floor is "h"
now here we can say that in vertical direction we have
initial speed is
[tex]v_y = 0[/tex]
acceleration is given as
[tex]a = g[/tex]
now we can use kinematics
[tex]y = v_y t + \frac{1}{2}at^2[/tex]
now plug in all values
[tex]h = 0 + \frac{1}{2}gt^2[/tex]
now by solving above equation we have
[tex]t = \sqrt{\frac{2h}{g}}[/tex]
So here it only depends on the height so in order to increase the time we need to increase the height.
