Question

A flea jumps straight up to a maximum height of 0.550 m . what is its initial velocity v0 as it leaves the ground?

Answer 1

To find the initial velocity of the flea that jumps to a height of 0.550 m, kinematic equations are used. Assuming the final velocity at the peak height is 0 m/s and the acceleration is -9.81 m/s^2 due to gravity, the calculated initial velocity is approximately 3.29 m/s.

Step-by-step explanation:

To calculate the initial velocity of the flea as it leaves the ground, we can use the principles of kinematics. More specifically, the kinematic equation for an object under constant acceleration due to gravity is:

v2 = u2 + 2as, where 'v' is the final velocity, 'u' is the initial velocity, 's' is the displacement, and 'a' is the acceleration (which, in this case, is the acceleration due to gravity, g = -9.81 m/s2).

When the flea reaches its maximum height, the final velocity ('v') is 0 m/s. Substituting 'v' with 0, 'g' with -9.81 m/s2, and 's' with the maximum height of 0.550 m, we get:

0 = u2 + 2(-9.81)(0.550)

After solving this equation for 'u', we find that the initial velocity of the flea as it leaves the ground is approximately 3.29 m/s.

Answer 2

Since my givens are x = .550m [Vsub0] = unknown
[Asubx] = =9.80

[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]

[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0])

Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in

0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)

0 = [Vsub0x]^2 -10.78

10.78 = [Vsub0x]^2

Sqrt(10.78) = 3.28 m/s