A golf ball is hit from the ground with an initial velocity at 30.0 degree above?

Question

A golf ball is hit from the ground with an initial velocity at 30.0 degree above the horizontal at t = 0.00s. At t =1.50s, it has a speed of 22.0m/s. calculate initial speed of the ball.

Answer 1

v=u+at

Where v is the final speed, u the initial speed and t is the time.

22=u+(-9.8)(1.50)
22=u-14.70
u=22+14.70
=36.70m/s

Answer 2

I believe with this problem, the 30 degrees doesn't matter...because distance isn't brought into play...you can use the equation:

V(f) - V(i) = ta

V(f) = final velocity
V(i) = initial velocity
t = change in time
a = acceleration due to gravity

V(i) = V(f) - ta
V(i) = (22m/s) - (1.50s)(9.8m/ss)
V(i) = 7.3m/s

so the initial velocity is 7.3m/s