Froghopper insects have a typical mass of around 11.7 mg and can jump to a heigh

Question

Froghopper insects have a typical mass of around 11.7 mg and can jump to a height of 49.3 cm. The takeoff velocity is achieved as the insect flexes its legs over a distance of approximately 2.00 mm. Assume that the jump is vertical and that the froghopper undergoes constant acceleration while its feet are in contact with the ground. Ignore air resistance. What is the acceleration of the insect during the time of the jump (before it leaves the ground)? m/s^2 During the jump, how long are the froghopper's feet in contact with the ground? What force did the ground exert on the froghopper during the jump? Express your answer as a multiple of the insect's weight.

Explanation / Answer

here,

mass , m = 11.7 mg = 1.17 * 10^-5 kg

height , h = 0.493 m

the final speed before the launch , v = sqrt(2 * g * h)

v = sqrt(2 * 9.81 * 0.493) m/s

v = 3.11 m/s

s = 2mm = 0.002 m

let the accelration be a

v^2 - u^2 = 2 * a * s

3.11^2 - 0 = 2 * a * 0.002

a = 2418.2 m/s^2

let the time taken for the jump be t

v = 0 + a * t

3.11 = 0 + 2418.2 * t

t = 1.28 ms

the force required , F = m * a

F = 1.17 * 10^-5 * 2418.2 N

F = 0.028 N

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