a) y(max)  = 337.76 m
b) t₁ = 5.30 s  the time for y maximum
c)t₂ =  13.60 s  time for y = 0 time when the fly finish
d) vₓ = 30 m/s        vy = - 81.32 m/s
e)x = 408 m
Equations for projectile motion:
v₀ₓ = v₀ * cosα          v₀ₓ = 60*(1/2)     v₀ₓ = 30 m/s   ( constant )
v₀y = v₀ * sinα           v₀y = 60*(√3/2)     v₀y = 30*√3  m/s
a) Maximum height:
The following equation describes the motion in y coordinates
y  =  y₀ + v₀y*t - (1/2)*g*t²      (1)
To find h(max), we need to calculate t₁ ( time for h maximum)
we take derivative on both sides of the equation
dy/dt  = v₀y  - g*t
dy/dt  = 0           v₀y  - g*t₁  = 0    t₁ = v₀y/g
v₀y = 60*sin60°  = 60*√3/2  = 30*√3
g = 9.8 m/s²
t₁ = 5.30 s  the time for y maximum
And y maximum is obtained from the substitution of t₁  in equation (1) 
y (max) = 200 + 30*√3 * (5.30)  - (1/2)*9.8*(5.3)²
y (max) = 200 + 275.40 - 137.64
y(max)  = 337.76 m
Total time of flying (t₂)  is when coordinate y = 0
y = 0 = y₀  + v₀y*t₂ - (1/2)* g*t₂²
0 = 200 + 30*√3*t₂  - 4.9*t₂²            4.9 t₂² - 51.96*t₂ - 200 = 0
The above equation is a second-degree equation, solving for  t₂
t =  [51.96 ±√ (51.96)² + 4*4.9*200]/9.8
t =  [51.96 ±√2700 + 3920]/9.8
t =  [51.96 ± 81.36]/9.8
t = 51.96 - 81.36)/9.8         we dismiss this solution ( negative time)
t₂ =  13.60 s  time for y = 0 time when the fly finish
The components of the velocity just before striking the ground are:
vₓ = v₀ *cos60°       vₓ = 30 m/s   as we said before v₀ₓ is constant
vy = v₀y - g *t        vy = 30*√3  - 9.8 * (13.60)
vy = 51.96 - 133.28          vy = - 81.32 m/s
The sign minus means that vy  change direction
Finally the horizontal distance is:
x = vₓ * t
x = 30 * 13.60  m
x = 408 m