Ball leaves the table of height 0.60 m
Initial speed of the ball is in horizontal direction which is 2.4 m/s
Now here we can say that time taken by the ball to hit the ground will be given by kinematics equation in vertical direction
[tex]\Delta y = v_y*t + \frac{1}{2}at^2[/tex]
here in y direction we can say
[tex]\Delta y = 0.60 m[/tex]
[tex]v_y = 0[/tex]
[tex]a = 9.8 m/s^2[/tex] due to gravity
now from above equation we have
[tex]0.60 = 0 + \frac{1}{2}*9.8*t^2[/tex]
[tex]t = 0.35 s[/tex]
Now in the same time the distance that ball move in horizontal direction is given as
[tex]d = v_x* t[/tex]
given that
[tex]v_x = 2.4 m/s[/tex]
[tex]d = 2.4 * 0.35[/tex]
[tex]d = 0.84 m[/tex]
So it will move a total distance of 0.84 m where it will land
