
4.25. Reconsider Example 4.14 and assume that the projectile speed and angle are correlated with
ρ
V
0
a
= 0.5. Compare the results with those of the uncorrelated case. Draw conclusions.
Solution: Recall
R =
υ
2
g
sin 2φ
µ
φ
= 30
◦
=
π
6
δ
φ
= 0.10
=⇒ σ
φ
= µ
φ
δ
φ
=
π
6
(0.10)
µ
υ
= 300 m/s
σ
υ
= 35 m/s.
For the general case of Y = g(X
1
, X
2
) where in the Taylor expansion terms to second-derivative are retained,
we have
E {Y } = g
µ
X
1
, µ
X
2
+
1
2
∂
2
g
∂X
2
1
V arX
1
+
∂
2
g
∂X
1
∂X
2
ρ
12
σ
X
1
σ
X
2
+
1
2
∂
2
g
∂X
2
2
V arX
2
V arY =
∂g
∂X
1
2
V arX
1
+ 2
∂g
∂X
1
∂g
∂X
2
ρ
12
σ
X
1
σ
X
2
+
∂g
∂X
2
2
V arX
2
,
where all derivatives are evaluated at the respective mean values. Therefore,
E(R) =
µ
2
υ
g
sin 2µ
φ
+
1
2
2
g
sin 2µ
φ
·σ
2
υ
+
4µ
υ
g
cos 2µ
φ
· ρ
φυ
σ
φ
σ
υ
+
1