Asked by Thea Foo
1. A balloon starts at position (1.3, -1.1, 3.1), where the units are in kilometers. If it is blown by the wind 16 kilometers along the two angles theta = 175 degrees and phi = -11 degrees, what is the distance from the balloon to the origin? How high is the balloon off the ground?
Answers
Answered by
Damon
in x y plane the balloon moves 16 cos 11 km = 15.7 km
so x displacement = 15.7 cos 175 = - 15.6
so final x= 1.3 -15.6 = 14.3
y displacement = 15.7 sin 175 = 1.37
so final y = -1.1 + 1.37 = 0.27
z displacement = -16 sin 11 = -3.05
so final z = 3.1 - 3.05 = 0.05 (about 50 meters above crash)
r = sqrt (14.3^2 + 0.27^2 + 0.05^2)
so x displacement = 15.7 cos 175 = - 15.6
so final x= 1.3 -15.6 = 14.3
y displacement = 15.7 sin 175 = 1.37
so final y = -1.1 + 1.37 = 0.27
z displacement = -16 sin 11 = -3.05
so final z = 3.1 - 3.05 = 0.05 (about 50 meters above crash)
r = sqrt (14.3^2 + 0.27^2 + 0.05^2)
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