Question

Two opposite walls of a deep gorge are sections of parallel vertical planes. One calm day, two ALLEN students Ram and Shyam positioned themselves at the edges of different walls of the gorge (see Fig. 2). The line connecting Ram and Shyam forms an angle $\phi$ = 30° with the horizontal. Ram and Shyam threw identical balls with the same initial speed $v_0$ relative to the Earth, in such a way that none of the thrown balls reached the opposite wall. The trajectories of the balls always lie in the vertical plane containing Ram and Shyam. A force of air resistance acts on a moving ball, directed opposite to the velocity of the ball and directly proportional to it ($\overrightarrow{F}$ = -k$\overrightarrow{v}$, where k is a constant). The sizes of Ram and Shyam can be neglected.

After conducting numerous experiments, Ram and Shyam discovered the following:

• If the ball was thrown horizontally, then it moved away horizontally at a distance S from the thrower (see Fig. 1);
• During simultaneous throws by Ram and Shyam such that Ram throws the ball horizontally with the speed $v_0$ and Shyam throws the ball with the speed $v_0$ with an angle $\alpha$ = 60° with the horizontal (see Fig. 2), the balls moved along the same vertical straight line after a very long time. (Take $v_0$ = 10 m/s, k = 2 N-s/m, m = 1kg)

The value of L is _____ m.

Answer 1

2.9

Answer 2

The problem involves analyzing projectile motion with air resistance. Key steps include:

1. Understanding the Forces: Identify the forces acting on the ball: gravity and air resistance. The air resistance is proportional to the velocity, $\overrightarrow{F} = -k\overrightarrow{v}$.

2. Equations of Motion: Set up the equations of motion in the horizontal and vertical directions, considering air resistance.

3. Horizontal Motion: The horizontal motion is governed by $\frac{dv_x}{dt} = -\gamma v_x$, where $\gamma = \frac{k}{m}$. Integrating this gives $x(t) = \frac{v_{x0}}{\gamma}(1 - e^{-\gamma t})$.

4. Vertical Motion: The vertical motion is governed by $\frac{dv_y}{dt} = -g - \gamma v_y$.

5. Asymptotic Horizontal Distance (S): When the ball is thrown horizontally, the maximum horizontal distance is $S = \frac{v_0}{\gamma} = \frac{10}{2} = 5$ m.

6. Relating Positions: Ram is at the origin, and Shyam is at $(L\cos\phi, L\sin\phi)$. The condition that the balls end up on the same vertical line after a long time means $x_1(\infty) = x_2(\infty)$.

7. Solving for L:

   • $x_1(\infty) = \frac{v_0}{\gamma}$
   • $x_2(\infty) = L\cos\phi + \frac{v_0\cos\alpha}{\gamma}$

   Equating the two: $\frac{v_0}{\gamma} = L\cos\phi + \frac{v_0\cos\alpha}{\gamma}$ $L\cos\phi = \frac{v_0}{\gamma}(1 - \cos\alpha)$

   Substituting the given values: $L\cos 30^\circ = \frac{10}{2}(1 - \cos 60^\circ) = 5(1 - 0.5) = 2.5$ $L\frac{\sqrt{3}}{2} = 2.5$ $L = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3} \approx 2.887$ m.

   Rounding to one decimal place, $L \approx 2.9$ m.