When jumping, a flea accelerates at an astounding 1400 m/s2, but over only the very short distance of 0.55 mm. If a flea jumps straight up, and if air resistance is neglected (a rather poor approximation in this situation), how high does the flea go

Given:

Acceleration, a = 1400 m/s^2

Distance, s = 0.55 mm

= 0.000055 m

Final velocity, v = 0 m/s

Using equation of motion,

v^2 = u^2 + 2as

u = √ (2as)

= √ (2 * 1400 * 0.000055)

= 0.8775 m/s

Initial velocity, u = 0.8775 m/s

Finding distance, S at, a = g = 9.8 m/s^2

Using equations of motion above,

S = u²/2g

= (0.8775) ² / (2 * 9.8)

= 0.039 m

= 39 mm

0.079 m or 79 mm

Explanation:

Using the equation of motion

v = √ (2as)

Where v is the velocity

a is acceleration = 1400m/s²

s is the distance = 0.55 mm = 0.00055m

Therefore

= √ (2 * 1400m/s² * 0.00055 m) = 1.54 m/s

Therefore; initial velocity = 1.54 m/s

Then we use the equation of motion s = v² / 2g

Take g = 9.8 m/s²

Therefore

= (1.54m/s) ² / 19.6 m/s²

= 0.079 m or 79 mm