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Inverse Trigonometric Functions

Question

A particle is projected from a point O with velocity u at an angle of 60^o with the horizontal. When it is moving in a direction at right angles to its direction at O, its velocity then is given by

u/3

u/2

2u/3

$fraction numerator u over denominator square root of 3 end fraction$

Solution

fraction numerator u over denominator square root of 3 end fraction

u cos 60^o = v cos 30^o

straight v equals space fraction numerator 4 over denominator square root of 3 end fraction