Wee can use here kinematics
as we know that
[tex]y = v*t + \frac{1}{2} at^2[/tex]
for shorter tree we know that
[tex]y = 0 + \frac{1}{2}*9.8 * 2^2[/tex]
[tex]y = 19.6 meter[/tex]
now since we know that other tree is twice high
So height of other tree is y = 39.2 m
now again by above equation
[tex]y = v*t + \frac{1}{2} at^2[/tex]
[tex]39.2 = 0 + \frac{1}{2}*9.8 * t^2[/tex]
[tex]t = 2.83 s[/tex]
so the time taken is 2.83 s