Two arrows are fired horizontally with the same speed of 30.0 m/s. each arrow has a mass of 0.100kg. one is fired due east and the other due south. find the magnitude and direction of the total momentum of this two-arrow system. specify the direction with respect to due east.

|momentum| = (mass) x (speed)

Momentum of each arrow = (0.1 kg) x (30 m/s) = 3 kg-m/s

Everything is equal for the two arrows, except their direction.
So the total momentum vector of the 2-arrow system consists of

-- a 3 kg-m/s component pointing east
plus
-- a 3 kg-m/s component pointing south.

The magnitude of the total momentum is

                √ [  (3 kg-m/s)²  +  (3 kg-m/s)²  ]

            =  √ [  9 kg²-m²/s²  +  9 kg²-m²/s²  ]

            =  √ [  18 kg²-m²/s²  ]

            =  3√2  kg-m/s        (approx  4.243 kg-m/s)

The direction of the total momentum vector is the angle
south of east whose tangent is

                             (3 kg-m/s) / (3 kg-m/s)  

                   =  tan⁻¹ (1)

                   =   45° south of east.

That direction is southeast.  

Ireally just need to get finished with precal! i’m really sorry i have to do this! hope this !