Physics problem: how to do, thanks?

(a) (i) x = V*cosΘ*t = 20m/s * cos30º * 1.2s = 21 m
y = V*sinΘ*t + ½at² = 20m/s*sin30º*1.2s - ½*9.8m/s²*(1.2s)²
y = 4.9 m

ii) Vx = 20m/s*cos30º = 17 m/s
Vy = 20m/s*sin30º - 9.8m/s²*1.2s = -1.8 m/s
|V| = √(Vx² + Vy²) = 17.4 m/s ≈ 17 m/s
Θ = arctan(Vy/Vx) = -5.8º = 5.8º below horizontal

(b) Assuming the ball is launched from ground level,
y = v*sinΘ*t + ½at²
so here 4 = 10t - 4.9t²
quadratic with solutions at t = 0.55 s and t = 1.5 s

(c) Still 4 = 10t - 4.9t²
and so the answers are the same

Hope this helps!

Vxo = 20cos30 = 17.32
Vyo = 20sin30 = 10
(a) 
(i) x = Vxo *t = 17.32*1.2 = 20.8
y = Vyo*1.2 - 4.9*t² = 12 - 4.9*1.2² = 4.944
Vx(1.2) = 17.32
(a)
(ii) Vy(1.2) = 10 - 9.8*1.2 = -17.6m/s 
17.6²+17.32² = 24.7² at arctan(-17.6/17.32) = -45°
4 = Vyo*t - 4.9t² = 10t-4.9t² at t = 0.5462, 1.4947
using the quadratic formula.