Question

a toy airplane ejects its 8.4 g pilot vertically using a spring with a spring constant of 5.2 x 10² n/m. if the spring is initially compressed by 5.2 cm down,
a) calculate the speed of the pilot as it ejects upward from the airplane.
b) determine the maximum height that the pilot will reach.

Explanation:

Step1: Convert units to SI

Mass: $m = 8.4\ \text{g} = 0.0084\ \text{kg}$
Spring compression: $\Delta x = 5.2\ \text{cm} = 0.052\ \text{m}$
Spring constant: $k = 5.2 \times 10^2\ \text{N/m}$

Step2: Use energy conservation for speed

Elastic potential energy converts to kinetic energy:
$$\frac{1}{2}k\Delta x^2 = \frac{1}{2}mv^2$$
Cancel $\frac{1}{2}$ and solve for $v$:
$$v = \sqrt{\frac{k\Delta x^2}{m}} = \Delta x\sqrt{\frac{k}{m}}$$
Substitute values:
$$v = 0.052\ \text{m} \times \sqrt{\frac{520\ \text{N/m}}{0.0084\ \text{kg}}}$$
$$v = 0.052 \times \sqrt{61904.76} \approx 0.052 \times 248.81$$

Step3: Calculate ejection speed

$$v \approx 12.94\ \text{m/s}$$

Step4: Find max height (part b)

Kinetic energy converts to gravitational potential energy:
$$\frac{1}{2}mv^2 = mgh$$
Cancel $m$ and solve for $h$:
$$h = \frac{v^2}{2g}$$
Use $g = 9.8\ \text{m/s}^2$ and $v \approx 12.94\ \text{m/s}$:
$$h = \frac{(12.94)^2}{2 \times 9.8} = \frac{167.44}{19.6}$$

Step5: Compute maximum height

$$h \approx 8.54\ \text{m}$$

Answer:

a) The speed of the pilot as it ejects is approximately $\boldsymbol{12.9\ \text{m/s}}$ (rounded to 2 significant figures, or $\boldsymbol{13\ \text{m/s}}$)
b) The maximum height the pilot reaches is approximately $\boldsymbol{8.5\ \text{m}}$ (rounded to 2 significant figures, or $\boldsymbol{8.5\ \text{m}}$)