Height to be reached above is given as
H = 25 m
let say initial speed is "v"
the final speed that will reach must be ZERO
now we will use kinematics
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
here we know that
[tex]v_f = 0[/tex]
[tex]v_i = ?[/tex]
[tex]a = -9.81 m/s^2[/tex]
[tex]d = 25 m[/tex]
now from above equation we will have
[tex]0 - v^2 = 2(-9.81)(25)[/tex]
[tex]v^2 = 490.5 m^2/s^2[/tex]
[tex]v = 22.15 m/s[/tex]
so the speed must be 22.15 m/s