Answer:
v_b = 10.628 m /s (13.605 degrees east of south)
Explanation:
Given:
- Velocity of mia v_m = + 6 j m/s
- Velocity of ball wrt mia v_bm = 5.0 m/s 30 degree due east of south
Find:
What are the magnitude and direction of the velocity of the ball relative to the ground? v_b
Solution:
- The relation of velocity in two different frame is given:
v_b - v_m = v_bm
- Components along the direction of v_b,m:
v_b*cos(Q) - v_m*cos(30) = 5
v_b*cos(Q) = 5 + 6 sqrt(3) / 2
v_b*cos(Q) = 5 + 3sqrt(3)
- Components orthogonal the direction of v_b,m:
-v_m*sin(30) = v_b*sin(Q)
-6*0.5 = v_b*sin(Q)
-3 = v_b*sin(Q)
- Divide two equations:
tan(Q) = - 3 / 5 + 3sqrt(3)
Q = arctan(- 3 / 5 + 3sqrt(3)
Q = -16.395 degrees
v_b = -3 / sin(-16.395)
v_b = 10.63 m/s
