A flea jumps from a hard surface reaching a maximum height of 10cm and returning

Question

A flea jumps from a hard surface reaching a maximum height of 10cm and returning to the surface 30cm away. The jump phase, when the flea is pushing against the surface covers a distance of 0.5mm.

Calculate the velocity with which the flea leaves the ground (vert and horiz components).

Calculate the average acceleration during the jump phase.

If the surface from which the flea jumps is moving at .2 m/s in the opposite direction of the flea's horizontal motion, what is the flea's initial velocity in the frame where the surface is moving?

Explanation / Answer

a)
For vertical motion:
Vfv^2 = Viv^2 - 2*g*h
0 = Viv^2 - 2*9.8*0.1
Viv = 1.4 m/s

time taken t = Viv/g = 1.4/9.8 = 0.143 s

for horizontal motion:
vih = horizontal distance/ time
= 0.3/0.143
=2.1 m/s
Answer:
Vertical component = 1.4 m/s
Horizontal component = 2.1 m/s

b)
for jump phase:
V = sqrt (1.4^2 + 2.1^2) = 2.52 m/s

a = v^2/2*d
= (2.52)^2 / (2*0.5*10^-3)
=6370 m/s^2

c)
In frame of reference:
Vertical component = 1.4 m/s
Horizontal component = 2.1 m/s - 0.2 m/s = 1.9 m/s

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