Here Sarah projected the ball upwards by speed
[tex]v_i = 19.62 m/s[/tex]
here the acceleration is due to gravity
[tex]a = 9.81 m/s^2[/tex]
now when ball again lends on the ground the displacement of ball will be zero
[tex]\delta y = v_y * t + \frac{1}{2} at^2[/tex]
[tex]0 = 19.62 t - \frac{1}{2}9.81*t^2[/tex]
by solving above equation
[tex]t = 4 s[/tex]
so it will take 4 s to reach back on ground
