Explanation:
Given that,
The x-component is 8 m/s east and y-component is 14.5 m/s North.
(a) As both x and y components are perpendicular to each other. Their resultant is given by :
[tex]R=\sqrt{x^2+y^2} \\\\R=\sqrt{(8)^2+(14.5)^2} \\\\R=16.56\ m/s[/tex]
(b) Let [tex]\theta[/tex] is the resultant vector. So,
[tex]\tan\theta=\dfrac{y}{x}\\\\\theta=\tan^{-1}(\dfrac{y}{x})\\\\\theta=\tan^{-1}(\dfrac{14.5}{8})\\\\\theta=61.11^{\circ}[/tex]
Hence, the resultant is 16.56 m/s and the angle of the resultant vector is 61.11 degrees.
