Question

Froghopper insects have a typical mass of around 12.5 mg and can jump to a height of 42.3 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)?

Answer 1

2065.005 m/s²

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

`v^2-u^2=2as\\\Rightarrow a=(v^2-u^2)/(2s)\\\Rightarrow a=(v^2-0^2)/(2* h)\\\Rightarrow v^2=2ah\ m/s`

Now H-h = 42.3 - 0.2 = 42.1 cm = 0.421 m

The final velocity will be the initial velocity

`v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2* g(H-h)\\\Rightarrow -2a0.002=2* g0.421\\\Rightarrow a=-(0.421* -9.81)/(0.002)\\\Rightarrow a=2065.005\ m/s^2`

Acceleration of the frog is 2065.005 m/s²