Jumping Flea - Simple Kinematics

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To determine the initial velocity of a flea that jumps to a maximum height of 0.53 m, the kinematic equation used is v_f^2 = v_i^2 + 2a_s(s_f - s_i). Given that the final velocity (v_f) at the peak height is 0 m/s, the acceleration due to gravity (a_s) is -9.8 m/s², the initial position (s_i) is 0 m, and the final position (s_f) is 0.53 m, the equation simplifies to 0 = v_i^2 - 2(9.8)(0.53). Solving for v_i, the initial velocity is found to be the positive root of 2(9.8)(0.53), confirming the calculation is correct. The analysis effectively demonstrates the application of kinematic equations in solving for initial velocity without time data.
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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: {v_f}^2 = {v_i}^2 + 2a_s(s_f - s_i). 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 a_s = -9.8. Also, s_i = 0 and s_f = 0.53.

Thus I have 0 = {v_i}^2 - 2(9.8)(0.53) \Leftrightarrow \sqrt[]{2(9.8)(0.53)} = v_i 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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