Vector is in the direction 34.0o clockwise from the –y-axis. The x-component of is Ax = - 16.0 m. (a) What is the y -component of ? (b) What is the magnitude of ?

Solution 32E Introduction. If A is making an angle with positive x axis, then, the x component of the vector is given by A x Acos ………….(1) And the y component is given by A y Asin ………….(2) Where A is the magnitude of the vector A , that is A = |A|. By dividing equation (2) with equation (1), we can get y tan = x ……………..(3) So if we know the x component A and the angle we can calculate the y component A of the vector. x y Now knowing x and y component of the vector we can easily calculate the magnitude of the vector by the formula A = Ax2+ A y Step 1 a) The condition given in the problem is shown in the following figure. So, from the figure we can write that the angle made by the vector with the positive x axis si = 180° 34° = 146° Hence from the equation (3) we can write that the y component is A y A xan = ( 16.0 m)tan146° = 10.8 m Hence the y component of the vector is 10.8 m.