Questions: In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80 m / s^2. Ignore air resistance. A flea jumps straight up to a maximum height of 0.410 m. What is its initial velocity v0 as it leaves the ground? Express your answer in meters per second to three significant figures.

Solution Steps

• Maximum height (\(h\)) = 0.410 m
• Acceleration due to gravity (\(g\)) = \(9.80 \, \mathrm{m/s^2}\)
• Final velocity at maximum height (\(v\)) = 0 m/s (since the flea momentarily stops at the peak)

Use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement: \[ v^2 = v_{0}^2 + 2a s \] where:

• \(v\) is the final velocity,
• \(v_{0}\) is the initial velocity,
• \(a\) is the acceleration,
• \(s\) is the displacement.

Substitute \(v = 0 \, \mathrm{m/s}\), \(a = -9.80 \, \mathrm{m/s^2}\) (negative because gravity acts downward), and \(s = 0.410 \, \mathrm{m}\): \[ 0 = v_{0}^2 + 2(-9.80)(0.410) \]

Rearrange the equation to solve for \(v_{0}^2\): \[ v_{0}^2 = -2(-9.80)(0.410) \] \[ v_{0}^2 = 2 \times 9.80 \times 0.410 \] \[ v_{0}^2 = 8.036 \]

Take the square root to find \(v_{0}\): \[ v_{0} = \sqrt{8.036} \] \[ v_{0} \approx 2.835 \, \mathrm{m/s} \]

Final Answer