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.

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.

Transcript text: In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be $9.80 \mathrm{~m} / \mathrm{s}^{2}$. Ignore air resistance. A flea jumps straight up to a maximum height of 0.410 m . What is its initial velocity $v_{0}$ as it leaves the ground? Express your answer in meters per second to three significant figures. View Available Hint(s) \[ v_{0}= \] $\square$

Solution

Solution Steps

Step 1: Identify the Known Values

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)

Step 2: Choose the Appropriate Kinematic Equation

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.

Step 3: Substitute the Known Values into the Equation

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

Step 4: Solve for the Initial Velocity $v_{0}$

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 $v_{0}$: \[ v_{0} = \sqrt{8.036} \] \[ v_{0} \approx 2.835 \, \mathrm{m/s} \]