Question

required information
a locust jumps at an angle of 55.0° and lands 0.830 m from where it jumped.
if it jumps with the same initial speed at an angle of 45.0°, the range would be
multiple choice
○ larger than the initial range.
○ two times smaller than the initial range.
○ the same as the initial range.
○ slightly smaller than the initial range.

Explanation:

Step1: Recall projectile range formula

The range of a projectile is given by $$R = \frac{v_0^2 \sin(2\theta)}{g}$$ where $v_0$ is initial speed, $\theta$ is launch angle, and $g$ is gravitational acceleration.

Step2: Relate the two ranges

Let $R_1$ be range at $\theta_1=55.0^\circ$, $R_2$ at $\theta_2=45.0^\circ$. Since $v_0$ and $g$ are constant, $\frac{R_2}{R_1} = \frac{\sin(2\theta_2)}{\sin(2\theta_1)}$.

Step3: Calculate the sine values

$2\theta_1 = 110.0^\circ$, $\sin(110.0^\circ) \approx 0.9397$; $2\theta_2 = 90.0^\circ$, $\sin(90.0^\circ) = 1.000$.

Step4: Compute the ratio

$\frac{R_2}{R_1} = \frac{1.000}{0.9397} \approx 1.064$, so $R_2 \approx 1.064 R_1$.

Answer:

A. larger than the initial range.