A jogger ran 3 miles due east of his house. then he ran 5 miles at a heading of 30o east of north (or 30o ne). how far is he from his house after running 8 miles

Magnitude of the net displacement: 7.0 milesStep-by-step explanation:In order to find the total displacement of the man, we have to resolve each displacement along the x- and y- direction.Taking x as positive x direction and y as positive north direction:- The first motion is 3 miles due east, so[tex]A_x = 3 mi\\A_y = 0[/tex]- The second motion is 5 miles at [tex]30^{\circ}[/tex] east of north, so:[tex]B_x = 5 sin 30^{\circ}=2.5 mi\\B_y = 5 cos 30^{\circ}=4.3 mi[/tex]So the components of the net displacement are[tex]R_x = A_x + B_x = 3 + 2.5 = 5.5 mi\\R_y = A_y+B_y = 0 + 4.3 = 4.3 mi[/tex]And so the magnitude of the net displacement is[tex]d=\sqrt{R_x^2+R_y^2}=\sqrt{5.5^2+4.3^2}=7.0 mi[/tex]So, he is 7.0 miles far from the house.Learn more about distance and displacement:brainly.com/question/3969582#LearnwithBrainly