Somebody please help

  • #1
A helicopter holding a 70-kilogram package suspended from a rop 5.0 meters long accelerates upward at a rate of 5.2 m/s^2. neglect air resistence on the package. When the upward velocity of the helicopter is 30 meters per second, the rope is cut and the helicopter continues to accelerate upward at 5.2 m/s^2. determine the diestance between the helicopter and the package 2.0 seconds after the rope is cut.

somebody please help. i dont understand how to do this problem.
 

Answers and Replies

  • #2
116
0
Start by drawing a sketch of the situation at the instant after the rope is cut.

Label the helicopter's velocity and acceleration, the distance between the helicopter and the package, the package's velocity and the forces acting on the package.

Then decide what formulas or laws might be applicable.
 
  • #3
can you give us any equations because we have sketched it out but dont know what equations to use.
 
  • #4
116
0
For the helicopter, since acceleration, initial velocity and position are given, use a position equation.

For the package, the acceleration is g (down), and the same type of equation can be used.
 
  • #5
so "x=x_0+v_0t+(1/2)at^2"
this?
 
  • #6
116
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That's the one.
 
  • #7
help us fill out this equation. is it "x=0+(30)(2)+(1/2)(5.2)(2^2)" is this the equation and we are working to find t^2?
 
Last edited:
  • #8
so we did the math for the helicopter and the package and got (after 2 seconds) an increase of 70.4 for the helicopter and an increase of -14.22 for the package so in order to find the distance between them after 2 seconds we added 70.4 and 14.22 and got a total distance difference of 80.62m. is this correct?
 
  • #9
seriously people please, tell me if we have the right answer or not
 
  • #10
i feel i have followed the rules this time and have given my imput and now want to know if i came to the right solution so if you wouldnt mind letting me know if this solution is right, that would be just wonderful
 

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