Answer:
The initial arrow's velocity is 40,9 m/s at 11.9° from the horizontal
Explanation:
In order to find the inital velocity we need to determine its components and the angle that the arrow is launched at.
For horizontal component, we will have:
ν cos(θ)t = x ⇒ cos(θ) = x/νt
For vertical component we will have:
h= v sin(θ)t ₋ gt²÷ 2 ⇒ sin (θ) = h + gt²÷2/νt
From the two equations we got, after noting that the vertical displacement is 3m, we can calculate
tan(θ) = h +gt²÷2/νt/ x÷νt = h+ gt²÷2/x = 3+ 9.8.0.5²÷2/20 = 0.21125
Now we can calculate θ = tan⁻¹(0.21125) ≈ 11.9°
Now that we know the angle we can subtitute at any of the expressions for the two components of the velocity . Let's do this subsitution at the horizontal component:
ν cos(θ)t= x =ν = x/tcos(θ)= 20/ 0.5cos(11.9) ≈ 40.9 m/s.