Answer:
A. 2.83 m/s
B. 39.55 degrees north of east
Explanation:
The velocity of the ball relative to the ground is the sum vectors of velocity of the ball relative to Mia and the velocity of Mia relative to the ground
If we take north east as positive direction then
Velocity vector of the ball relative to Mia is
[tex]<3.6 sin30^0, -3.6cos 30^0 > = <3.6 * 1/2, -3.6 *0.866> = <1.8, -3.12>[/tex]
Velocity of Mia relative to gorund is
<0, 5.3>
So velocity of the ball relative to the ground is
<1.8, -3.12> + <0, 5.3> = <1.8, 2.18>
Its magnitude is:
[tex]v = \sqrt{v_e^2 + v_n^2} = \sqrt{1.8^2 + 2.18^2} = \sqrt{3.24 + 4.7524} = \sqrt{7.9924} = 2.83[/tex] m/s
Its direction is
[tex]tan\alpha = \frac{v_e}{v_n} = \frac{1.8}{2.18} = 0.83[/tex]
[tex]\alpha = tan^{-1}0.83 = 0.69 rad \approx 39.55 degrees [/tex] north of east
