Question

In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80m/s2 . Ignore air resistance. A flea jumps straight up to a maximum height of 0.460m . What is its initial velocity v0 as it leaves the ground?

Answer 1

xf = 0. xi = 0. (you start and end at the same position, the ground). Your initial velocity is 2.62. gravity is 4.905.

xf = xi + vi(t) + 1/2(a)t^2.

0 = 0 + 2.62(t) + 1/2(-9.81)t^2
0 = 2.62t - 4.905t^2
4.905t = 2.62
t = 2.62/4.905
t = 0.5341488277268093781855249745158
time in three significant figures:
t = 0.534.

Answer 2

As the flea jumps and leaves the ground, its speed is 'V₀'.

When the flea reaches its maximum height, its speed is zero.
(That's why it doesn't go any higher than that.)

Its average speed all the way up is

(1/2) (V₀ + 0) = V₀ / 2 .

The time it takes for the original speed to dribble down to zero is

V₀/g = V₀/9.8 .

Distance covered = (Average speed) x (time in motion)

0.46 m = (V₀/2) x (V₀/9.8)

0.46 = V₀² / 19.6

V₀² = (0.46 x 19.6) = 9.016

V₀ = √9.016 = 3 m/s (rounded)