Measuring Height to Calculate Froghopper Jump Height

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The froghopper (Philaenus spumarius) is recognized for its remarkable jumping ability, achieving an acceleration of 4400 m/s² over 2.0 mm. The calculated velocity after this acceleration is 4.19 m/s, taking approximately 0.952 ms to reach that speed. For part (b), to determine the jump height without air resistance, the appropriate equation to use is v² = v₀² + 2ay, where gravity will decelerate the insect after it leaves the ground. The discussion highlights the importance of understanding that once airborne, the froghopper can no longer accelerate and will decelerate due to gravity. This insight is crucial for accurately calculating the theoretical jump height.
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help with part b

It has been claimed that an insect called the froghopper (Philaenus spumarius) is the best jumper in the animal kingdom. This insect can accelerate at 4400 m/s2 over a distance of 2.0 mm as it straightens its specially designed "jumping legs."

I got this right

(a) Assuming a uniform acceleration, what is the velocity of the insect after it has accelerated through this short distance, and how long did it take to reach that velocity?
velocity 4.19 m/s
elapsed time .952 ms




(b) How high would the insect jump if air resistance could be ignored? Note that the actual height obtained is about 0.7 m, so air resistance is important here.

I don't know what to for part b

should I plug into the equation v^2= v0^2 + 2ax?

or what? Its driving me insane
 
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is the bug jumping verticaly?
 
Once the bug leaves the ground gravity takes over and he will start to delcelarate due to gravitational attraction. That is it cannot accelerate itself any more since it cannot push against the surface. So it will decelerate at a rate of g. Your choice of equation will therefore do just fine. But maybe you should replace the x with a y (joking).
 
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