Answer:
C. 590 mph
Explanation:
Given:
- velocity of jet,
- direction of velocity of jet, east relative to the ground
- velocity of Cessna,
- direction of velocity of Cessna, 60° north of west
Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.
Refer the attached schematic.
So,
&
Now the vector of relative velocity of Cessna with respect to jet:
Now the magnitude of this velocity:
is the relative velocity of Cessna with respect to the jet.
The force acting on his feet.
The horizontal force is m*v²/Lh, where m is the total mass. The vertical force is the total weight (233 + 840)N.
<span>Fx = [(233 + 840)/g]*v²/7.5 </span>
<span>v = 32.3*2*π*7.5/60 m/s = 25.37 m/s </span>
<span>The horizontal component of force from the cables is Th + Ti*sin40º and the vertical component of force from the cable is Ta*cos40º </span>
<span>Thh horizontal and vertical forces must balance each other. First the vertical components: </span>
<span>233 + 840 = Ti*cos40º </span>
<span>solve for Ti. (This is the answer to the part b) </span>
<span>Horizontally </span>
<span>[(233 + 840)/g]*v²/7.5 = Th + Ti*sin40º </span>
<span>Solve for Th </span>
<span>Th = [(233 + 840)/g]*v²/7.5 - Ti*sin40º </span>
<span>using v and Ti computed above.</span>
The answer is B
second law
