What was the highest point of the trajectory?

Question

A rock is thrown 60 degrees of the ground with a initial velocity of 7.22m/s.

Answer 1

Vertical velocity = 7.22 sin 60° = 6.25 m/s.
Find height via v^2 = u^2 + 2as
=> 0 = 6.25^2 + 2(-9.81)s
=> s = 1.99 m.

Answer 2

At highest point, Final Velocity = 0
o=Initial velocity ^ 2 + 2*A*S
Since against gravity, A=-9.8
now 0=Initial Velocity ^ 2 - 2*9.8*S
=> Initial Velocity ^ 2 = 2*9.8*S
7.22^2 = 19.6*S
=> S = 7.22^2/19.6
= 5.32
But this is when the trajectory is vertical. But it is thrown at an angle of 60 degrees. So the new displacement is
S = S/Tan (60)
S = 5.32/0.5
S = 10.64