Respuesta :
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
In this case, the magnitude and direction of the velocity of the ball relative to the ground - 3.01 and 56.13 degrees northeast.
The velocity of the ball relative to the ground is the sum vectors of the velocity of the ball relative to Mia and the velocity of Mia relative to the ground
therefore,
- assuming northeast as a positive direction then the velocity vector of the ball relative to Mia is:
[tex]<5sin30^{0}, -5scos30^{0}>\\\\[/tex]
[tex]<5*0.5, -5*0.866 >\\\\< 2.5, - 4.33 >[/tex]
- Velocity of Mia relative to gorund is
<0, 6>
So velocity of the ball relative to the ground is
<2.5, -4.33> + <0, 6> = <2.5, 1.67>
Its magnitude is:
[tex]v = \sqrt{v_{e}^{2} + v_{n}^{2} }[/tex]
= [tex]\sqrt{2.5^{2} + 1.67^{2} }\\[/tex]
[tex]\sqrt{6.25 + 2.79 }\\\\= \sqrt{9.04}\\\\= 3.01\\[/tex]
direction:
[tex]tan \alpha= \frac{v_e}{v_n}\\ = \frac{2.5}{1.67}\\ = 1.49\\\alpha = tan^{-1}1.49\\ = 56.13^{\circ}[/tex]nort east.
Thus, the magnitude and direction of the velocity of the ball relative to the ground - 3.01 and 56.13 degrees northeast.
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https://brainly.com/question/17170248
