When jumping, a flea reaches a takeoff speed of 1.2 m/s over a distance of 0.46m

Question

When jumping, a flea reaches a takeoff speed of 1.2 m/s over a distance of 0.46mm.

With answers use two significant figures.

A)What is the flea's acceleration during the jump phase?

B)How long does the acceleration phase last?

C)If the flea jumps straight up, how high will it go? (Ignore air resistance for this problem; in reality, air resistance plays a large role, and the flea will not reach this height.)

Please help with showing all the steps so I can learn the process of solving questions like this.

Explanation / Answer

1. Use v^2 = v0^2 + 2*a*h and solve for "a".

v = 1.2 m/s
v0 = 0
h = 0.46 mm = 0.00046 m

a = 1.2*1.2 / 2*0.00046 = 1565.21

2. Now that you got a, use v = v0 + a*t and solve for the time. v and v0 are the same as before.

t = v-v0 / a

t = 1.2/1565.21 = 0.76 milli sec

3. For the third question you can assume that initial speed is the one reached during takeoff v0 = 1.3 m/s. Maximum height is reached when velocity becomes zero v = 0. The acceleration is the gravitational one g = 9.8 m/s^2. And the equation is

v^2 = v0^2 -2*g*hmax

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