1. **Problem statement:** Two bodies are thrown simultaneously from the same point. One is thrown straight up with initial velocity $v_0 = 20$ m/s, and the other is thrown at an angle $\theta = 30^\circ$ with the same initial velocity $v_0 = 20$ m/s. We need to find the distance between the two bodies at time $t = 1.2$ s, neglecting air resistance. 2. **Formulas and concepts:** - For projectile motion, the horizontal and vertical components of velocity are: $$v_{0x} = v_0 \cos \theta, \quad v_{0y} = v_0 \sin \theta$$ - The position at time $t$ is: $$x = v_{0x} t$$ $$y = v_{0y} t - \frac{1}{2} g t^2$$ - For the body thrown straight up, $\theta = 90^\circ$, so $v_{0x} = 0$ and $v_{0y} = v_0$. - Gravity acceleration $g = 9.8$ m/s$^2$. 3. **Calculate positions at $t=1.2$ s:** - Body 1 (straight up): $$x_1 = 0$$ $$y_1 = v_0 t - \frac{1}{2} g t^2 = 20 \times 1.2 - 0.5 \times 9.8 \times (1.2)^2$$ $$y_1 = 24 - 0.5 \times 9.8 \times 1.44 = 24 - 7.056 = 16.944 \text{ m}$$ - Body 2 (angle $30^\circ$): $$v_{0x} = 20 \cos 30^\circ = 20 \times \frac{\sqrt{3}}{2} = 17.32 \text{ m/s}$$ $$v_{0y} = 20 \sin 30^\circ = 20 \times 0.5 = 10 \text{ m/s}$$ $$x_2 = v_{0x} t = 17.32 \times 1.2 = 20.784 \text{ m}$$ $$y_2 = v_{0y} t - \frac{1}{2} g t^2 = 10 \times 1.2 - 0.5 \times 9.8 \times 1.44 = 12 - 7.056 = 4.944 \text{ m}$$ 4. **Calculate the distance between the two bodies:** $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(20.784 - 0)^2 + (4.944 - 16.944)^2}$$ $$= \sqrt{(20.784)^2 + (-12)^2} = \sqrt{431.87 + 144} = \sqrt{575.87} \approx 24.0 \text{ m}$$ 5. **Answer:** The distance between the bodies at $t=1.2$ s is approximately 24 m, which corresponds to option (B).