A 20 mm long, 0.50 mg flea can reach a height of 20 cm in a single leap. Ignoring air drag, what is the take-off speed of such a flea?

Steps to Solve

1. Identify the variables needed We need the height ($h$) the flea is jumping, which is given as $20 \text{ cm} = 0.20 \text{ m}$. We also need the acceleration due to gravity ($g$), which is approximately $9.81 \text{ m/s}^2$.

2. Use the conservation of energy principle To calculate the required take-off speed ($v_0$), we can use the principle of conservation of energy. At the highest point, all kinetic energy will be converted to potential energy.

The potential energy ($PE$) at height $h$ is given by: $$ PE = mgh $$ where $m$ is the mass of the flea.

3. Relate potential and kinetic energy The initial kinetic energy ($KE$) needed to reach this height can be expressed as: $$ KE = \frac{1}{2} mv_0^2 $$ At the maximum height, all kinetic energy is converted to potential energy: $$ \frac{1}{2} mv_0^2 = mgh $$

4. Simplify the equation The mass $m$ cancels out from both sides of the equation: $$ \frac{1}{2} v_0^2 = gh $$

5. Solve for the take-off speed Rearranging the equation gives: $$ v_0^2 = 2gh $$ We can find $v_0$ by taking the square root: $$ v_0 = \sqrt{2gh} $$

6. Plug in the values Substituting the known values into the equation: $$ v_0 = \sqrt{2 \times 9.81 , \text{m/s}^2 \times 0.20 , \text{m}} $$

7. Calculate the take-off speed Perform the calculation: $$ v_0 = \sqrt{2 \times 9.81 \times 0.2} = \sqrt{3.924} \approx 1.979 \text{ m/s} $$