Source: High School Physics Olympiad
Problem Sets:
Problem
An object is thrown with initial speed $v_0 = 20$ m/s at an elevation angle $\alpha = 60°$. Air resistance is neglected; take $g = 9.8$ m/s$^2$.
- At the end of 1.5 s after the motion begins, what is the angle $\theta$ between the direction of motion and the horizontal?
- How long after launch does the direction of motion make a 45° angle with the horizontal, and what is the height of the object at that moment?
- What are the radii of curvature of the trajectory at the highest point and at the landing point?
The velocity components are $v_x = v_0\cos\alpha = 10$ m/s (constant) and $v_y = v_0\sin\alpha - gt = 10\sqrt{3} - 9.8t$.
Q1: At $t = 1.5$ s: $v_y = 17.32 - 14.7 = 2.62$ m/s, so
$$\tan\theta = \frac{v_y}{v_x} = \frac{2.62}{10} = 0.262, \qquad \theta \approx 15°$$Q2: A 45° direction requires $v_y = \pm v_x = \pm 10$ m/s. Ascending: $10\sqrt{3} - 9.8t = 10$ gives $t = \dfrac{10(\sqrt{3} - 1)}{9.8} \approx 0.75$ s (descending: $t \approx 2.79$ s). The height follows from $v_y^2 = (v_0\sin\alpha)^2 - 2gy$:
$$y = \frac{300 - 100}{2(9.8)} \approx 10.2 \text{ m}$$Q3: The radius of curvature is $\rho = v^2/a_n$ where $a_n$ is the acceleration component perpendicular to the velocity.
At the highest point $v = v_x = 10$ m/s and $a_n = g$:
$$\rho_1 = \frac{v_x^2}{g} = \frac{100}{9.8} \approx 10.2 \text{ m}$$At the landing point $v = v_0 = 20$ m/s (same height as launch) and $a_n = g\cos\alpha$:
$$\rho_2 = \frac{v_0^2}{g\cos\alpha} = \frac{400}{9.8 \times 0.5} \approx 81.6 \text{ m}$$