Question

A flea jumps straight up to a maximum height of 0.450 m . What is its initial velocity v0 as it leaves the ground?

Answer 1

The initial velocity of a flea that jumps to a height of 0.450 m can be found using the kinematic equation for uniformly accelerated motion, yielding approximately 2.97 m/s.

Step-by-step explanation:

To determine the initial velocity (v₀) with which a flea jumps straight up to a maximum height of 0.450 m, we can use the physics principles of kinematics under the force of gravity. Since we are interested in the initial velocity, we'll use the formula derived from kinematic equations:

v² = u² + 2as, where:

• v is the final velocity (which is 0 m/s at the maximum height),
• u is the initial velocity,
• a is the acceleration (which is the acceleration due to gravity, g = -9.81 m/s², and it's negative because it's directed downward),
• s is the displacement (which is the maximum height, 0.450 m).

Rearranging the equation for u, we get:

u = √(v² - 2as)

Substituting the known values:

u = √(0 - 2(-9.81 m/s²)(0.450 m))

u = √(8.829 m/s)

Therefore, the initial velocity of the flea as it leaves the ground is approximately 2.97 m/s.

Answer 2

Initial velocity =
`\(v_0\)`

acceleration in the downward direction = -9.8
`\(\frac {m}{s^2}\)`

Final velocity at the highest point = 0

Maximum height reached = 0.450 m

Now, Using third equation of motion:

`\(v^2 = {v_0}^(2) + 2aH`

`\(0^2 = {v_0}^(2) - 2 * 9.8 * 0.450`

`\({v_0}^(2) = 2 * 9.8 * 0.450\)`

`\(v_0 = 2.969 \frac {m}{s}\)`

Velocity with which the flea jumps =
`\(2.969 \frac {m}{s}\)`