Physics problem?

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vertical motion equation 
hf-hi = Vo*sin 30°*t-g/2*t^2
(0-1)-1.5*0.5*t+4.903t^2 = 0 
4.903t^2-3t/4-1 = 0
time t = (0.75+√(0.75^2+4.903*4)/9.806 = 0.535 sec 

final speed V = √Vo^2+2gh = √1.5^2+19.612*1 = 4.68 m/sec (conservation  of energy)
or
final speed V = √(Vo*cos 30°)^2+(g*t-Vo*sin 30)^2
V = √1.5^2*3/4+(9.806*0.535-1.5*0.5)^2 = √1.688+4.50^2 = 4.68 m/sec 

s₀ = 1 m
u = 1.5 m/s
vertical component of velocity, uy = 1.5sin30°
θ = 30°
(a) When cork reaches the ground s = 0

s = s₀ + uy*t + ½gt²
0 = 1 + (1.5sin30)t + ½(-9.8)t²
4.9t² - 0.75t - 1 = 0

Using quadratic solution
t = {-(-0.75)±√[(-0.75)²-4(4.9)(-1)]}/(2*4.9)
t = 1.5 ±√[(-1.5)²-4(4.9)(-1)]}/(2*4.9)
t = 0.53 s

Cork reaches the ground in 0.53 s

(b) Let vertical component of velocity when cork reaches ground = vy
vy = uy + gt
vy = 1.5sin30 + (-9.8)(0.53)
vy = -4.4 (negative sign denotes cork travelling downwards)

horizontal component of initial velocity, ux = 1.5cos30, which will remain until the cork reaches the ground

resultant velocity =  √[(-4.4)² + (1.5cos30)²] = 4.6 m/s

speed of the cork when it reaches the ground is 4.6 m/s

(a) vertical kinematics:
y = y₀ + Vy₀*t + ½at²
0 = 1 + 1.5*sin30º*t - ½*9.8*t² = 1 + 0.75t - 4.9t²
quadratic with roots at
t = -0.38 s ← ignore negative root
and t = 0.53 s

(b) Easiest is v² = u² + 2gh = 1.5² + 2*9.8*1 = 22 (m²/s²)
v = 4.7 m/s