An object is launched up a 30-degree slope at an angle of 20 degrees above the incline of the slope and lands a distance L= 3.12 m up the slope. What is the object's initial speed?

An object is launched up a 30-degree slope at an angle of 20 degrees above the incline of the slope and lands a distance L= 3.12 m up the slope.  What is the object's initial speed?
ϴ₁=30°
ϴ₂=20°
L= 3.12 m
v₀=?
v₀ cos(ϴ₁+ϴ₂) ∙t =L∙cos(ϴ₁)
v₀ sin(ϴ₁+ϴ₂) ∙t - ½gt² = L∙sin(ϴ₁)
t ={L∙cos(ϴ₁) / (v₀ cos(ϴ₁+ϴ₂))}

v₀ sin(ϴ₁+ϴ₂) ∙{L∙cos(ϴ₁) / (v₀ cos(ϴ₁+ϴ₂))} - ½g{L∙cos(ϴ₁) / (v₀ cos(ϴ₁+ϴ₂))}² = L∙sin(ϴ₁)

sin(ϴ₁+ϴ₂) ∙L∙cos(ϴ₁) / cos(ϴ₁+ϴ₂) - ½gL²∙cos²(ϴ₁) / (v₀² cos²(ϴ₁+ϴ₂)) = L∙sin(ϴ₁)

sin(ϴ₁+ϴ₂) ∙cos(ϴ₁) / cos(ϴ₁+ϴ₂) - ½gL∙cos²(ϴ₁) / (v₀² cos²(ϴ₁+ϴ₂)) = sin(ϴ₁)

½gL∙cos²(ϴ₁) / (v₀² cos²(ϴ₁+ϴ₂))=sin(ϴ₁+ϴ₂) ∙cos(ϴ₁)/cos(ϴ₁+ϴ₂)-sin(ϴ₁)

½gL∙cos²(ϴ₁) / (v₀² cos²(ϴ₁+ϴ₂))=
=(sin(ϴ₁+ϴ₂) ∙cos(ϴ₁) -cos(ϴ₁+ϴ₂)∙sin(ϴ₁))/cos(ϴ₁+ϴ₂)= sin(ϴ₂)/cos(ϴ₁+ϴ₂)

½gL∙cos²(ϴ₁) / (v₀² cos²(ϴ₁+ϴ₂))=sin(ϴ₂)/cos(ϴ₁+ϴ₂)

v₀² cos²(ϴ₁+ϴ₂) / (½gL∙cos²(ϴ₁))   =  cos(ϴ₁+ϴ₂)/sin(ϴ₂)
v₀²  / cos²(ϴ₁)   =  gL/(2sin(ϴ₂)cos(ϴ₁+ϴ₂))

v₀/cos(ϴ₁)   =  √{gL/(2sin(ϴ₂)cos(ϴ₁+ϴ₂))}
v₀   =  cos(ϴ₁)√{gL/(2sin(ϴ₂)cos(ϴ₁+ϴ₂))}
Calculation:

cos(30deg)*sqrt(9.81m/s^2*3.12m/(2*sin(20deg)*cos(30deg+20deg)))=
=7.22549898 m/s
v₀ ≈ 7.23 m/s

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