It has been claimed that an insect called the froghopper (Philaenus spumarius) i

Question

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 3750 m/s2 over a distance of 1.7 mm as it straightens its specially designed "jumping legs."

(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 ???? m/2

elapsed time ???? 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.
??? m

Explanation / Answer

Given Acceleration of the insect a =  3750 m/s^2 The distance it can accelerate is , s = 1.7 mm =0.0017m a) v^2 - vi^2 = 2as    v ^2 = 2 *(3750 m/s^2) *0.0017 m    v = 3.5 m/s b) v = vi + at 3.5 m/s =3750 m/s^2 *t    t = 9.5*10 ^-4 s c) The height reached is H = v^2/ 2g     = (3.5 m/s)^2/ 2 *(9.8 m/s^2) =0.625 m Given Acceleration of the insect a =  3750 m/s^2 The distance it can accelerate is , s = 1.7 mm =0.0017m a) v^2 - vi^2 = 2as    v ^2 = 2 *(3750 m/s^2) *0.0017 m    v = 3.5 m/s b) v = vi + at 3.5 m/s =3750 m/s^2 *t    t = 9.5*10 ^-4 s c) The height reached is H = v^2/ 2g     = (3.5 m/s)^2/ 2 *(9.8 m/s^2) =0.625 m Given Acceleration of the insect a =  3750 m/s^2 The distance it can accelerate is , s = 1.7 mm =0.0017m a) v^2 - vi^2 = 2as    v ^2 = 2 *(3750 m/s^2) *0.0017 m    v = 3.5 m/s b) v = vi + at 3.5 m/s =3750 m/s^2 *t    t = 9.5*10 ^-4 s c) The height reached is H = v^2/ 2g     = (3.5 m/s)^2/ 2 *(9.8 m/s^2) =0.625 m

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