Jumping Flea - Simple Kinematics

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SUMMARY

The discussion focuses on calculating the initial velocity of a flea that jumps to a maximum height of 0.53 meters using kinematic equations. The relevant equation applied is {v_f}^2 = {v_i}^2 + 2a_s(s_f - s_i), where the final velocity (v_f) is 0 m/s at the peak height, the acceleration (a_s) is -9.8 m/s², and the initial (s_i) and final positions (s_f) are 0 m and 0.53 m, respectively. The solution confirms that the initial velocity (v_i) is calculated as v_i = √(2 * 9.8 * 0.53), yielding a positive root as the correct value.

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Homework Statement



A flea jumps straight up to a maximum height of 0.53 m. What is its initial velocity as it leaves the ground?


Homework Equations





The Attempt at a Solution



Okay, I have three kinematic equations for constant acceleration I can use. All of them require the knowledge of some time interval, except for one, so I shall use that one since I'm not given any time info in this question.

The equation says: [tex]{v_f}^2 = {v_i}^2 + 2a_s(s_f - s_i)[/tex]. Obviously f denotes final and i denotes initial.

I know that my final velocity will have to be 0 since if the flea jumps to a maximum height of 0.53 m, he will start coming down after that. I also know that [tex]a_s = -9.8[/tex]. Also, [tex]s_i = 0[/tex] and [tex]s_f = 0.53[/tex].

Thus I have [tex]0 = {v_i}^2 - 2(9.8)(0.53) \Leftrightarrow \sqrt[]{2(9.8)(0.53)} = v_i[/tex] and I will take the positive root to be my v_i.

Is this correct?
 
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Yes. It is correct.
 

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