Answer to Question #81066 in Classical Mechanics for Laurence

Question

A pilot flies with a heading of N 53o E, and its air speed is 154.8 miles/h. It is driven from its course by a wind of 396.4 miles/h and travelling S 14o E. Find the principle angle of the resultant.

Round the answer to 2 decimal places.

Answer 1

R_x=v_x+u_x=154.8 cos⁡53+396.4 cos⁡14=477.8 mph.
R_y=v_y+u_y=154.8 sin⁡53-396.4 sin⁡14=27.73 mph.
The principle angle of the resultant is
θ=tan^(-1)⁡〖R_y/R_x 〗=tan^(-1)⁡〖(27.73 )/477.8〗=N 3.32° E.