Answer:
a
x-component [tex]20 \ m[/tex]
y-component [tex]500 - 452 = 48 \ m[/tex]
b
Magnitude [tex]d = 52 \ m[/tex]
direction is [tex]\theta = 67.4^o[/tex]
Explanation:
From the question we are told that
The first vertical distance is [tex]y_1 = 500 \ m[/tex]
The first horizontal distance is [tex]x = 20 \ m[/tex]
The second vertical distance is [tex]y_2 = 452 \ m[/tex]
Generally the displacement is
x-component [tex]20 \ m[/tex]
y-component [tex]500 - 452 = 48 \ m[/tex]
Generally the helicopters displacement is mathematically evaluated as
[tex]d = \sqrt{ x- component ^2 + y- component ^2 }[/tex]
[tex]d = \sqrt{ 20t ^2 + 48 ^2 }[/tex]
[tex]d = 52 \ m[/tex]
The direction is the angle the displacement of the helicopter makes with the horizontal which is mathematically evaluated as
[tex]\theta = tan ^{-1}[ \frac{48}{20}][/tex]
=> [tex]\theta = tan ^{-1}[ 2.4 ][/tex]
=> [tex]\theta = 67.4^o[/tex]
