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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?

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MAXTORRES27
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. 
AL2006
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)

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