Question

required information
a locust jumps at an angle of 55.0° and lands 0.830 m from where it jumped.

what is the maximum height of the locust during its jump? ignore air resistance.
□ cm

Explanation:

Step1: Recall range formula

The range $R$ of projectile motion is given by:
$$R = \frac{v_0^2 \sin(2\theta)}{g}$$
where $R=0.830\ \text{m}$, $\theta=55.0^\circ$, $g=9.81\ \text{m/s}^2$.

Step2: Solve for initial velocity squared

Rearrange to solve for $v_0^2$:
$$v_0^2 = \frac{Rg}{\sin(2\theta)}$$
Substitute values:
$$v_0^2 = \frac{0.830 \times 9.81}{\sin(110.0^\circ)} = \frac{8.1423}{0.9397} \approx 8.665\ \text{m}^2/\text{s}^2$$

Step3: Recall max height formula

Maximum height $H$ is:
$$H = \frac{v_0^2 \sin^2(\theta)}{2g}$$

Step4: Substitute $v_0^2$ and calculate $H$

$$H = \frac{8.665 \times \sin^2(55.0^\circ)}{2 \times 9.81}$$
First, $\sin(55.0^\circ) \approx 0.8192$, so $\sin^2(55.0^\circ) \approx 0.6711$:
$$H = \frac{8.665 \times 0.6711}{19.62} \approx \frac{5.815}{19.62} \approx 0.2964\ \text{m}$$

Step5: Convert to centimeters

$$H = 0.2964 \times 100 = 29.64\ \text{cm}$$

Answer:

29.6 cm (rounded to three significant figures)