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In 1990, Gary Stewart of California made 177 737 jumps on a pogo stick. Suppose that the pogo stick reaches a height of 12.0 cm with each jump and that the average net force acting on the pogo stick during the contact with the ground is 330 N upward. What is the time of contact with the ground between the jumps? Assume the total mass of Stewart and the pogo stick is 65 kg. (Hint: The difference between the initial and final velocities is one of direction rather than magnitude.)

Respuesta :

The relationship between the impulse and the momentum allows to find the result for the question about the contact time is:

  • The contact time on the floor is t = 0.60 s

The impulse is the push force on a body and is related to the variation of the amount of momentum.

           I = ∫ F. dt = Δp

           p = mv

where I is the momentum, F the force, t the time, p the amount of momentum and m the mass .

They indicate the average force F = 330 N, the jump height is y = 12.0 cm = 0.120 m.

           

             F. t = m v_f - m v₀

Let's search with kinematics for the initial speed of the jump.

          v² = v₀² - 2 g y

At the highest point the velocity is zero.

           0 = v₀² - 2ay

           [tex]v_o = \sqrt{2gy}[/tex]  

Let's calculate.

         [tex]v_o = \sqrt{2 \ 9.8 \ 0.120}[/tex]  

         v₀ = 1.53 m / s

Suppose that the jumps are elastic, that is, that the speed with which it reaches the ground is equal to the speed with which it leaves, but in the opposite direction.

         [tex]t = \frac{m ( v_f - v_o)}{F} \\t =\frac{ m 2v_o }{F}[/tex]

   

Let's calculate.

         [tex]t= \frac{2 \ 65 \ 1.53}{330}[/tex]  

         t = 0.60 s

In conclusion using the relationship between impulse and momentum we can find the result for the question about the contact time is:  

         The contact time on the floor is t = 0.60 s

Learn more about impulse here: brainly.com/question/904448

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