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A 200 -lb object is released from rest 600ft above the ground and allowed to fall under the influence of gravity. Assuming that the force in pounds due to air resistance is -10v, where v is the velocity of the object in ft/sec, determine the equation of motion of the object. When will the object hit the ground?Assume that the acceleration due to gravity is 32 ft divided by sec squared32 ft/sec2 and let​ x(t) represent the distance the object has fallen in t seconds.

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Answer:

Step-by-step explanation:

  • Given mg = 200lb, g = 32ft/s^2
  • m = 200lb/32 = 6.25slug

Application of newton's second law ;

  • mdv/dt = mg - 10v, where g = 32ft/s^2
  • dv/dt = g - 10v/m
  • dv/dt = 32 - 1.6v

The detailed steps and appropriate integration is as shown in the attached file.

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