Question

A ball is thrown from the location \((x_0, y_0) = (0, 0)\) of a horizontal playground with an initial speed \(v_0\) at an angle \(\theta_0\) from the \(+x\)-direction. The ball is to be hit by a stone, which is thrown at the same time from the location \((x_1, y_1) = (L, 0)\). The stone is thrown at an angle \((180^\circ - \theta_1)\) from the \(+x\)-direction with a suitable initial speed. For a fixed \(v_0\), when \((\theta_0, \theta_1) = (45^\circ, 45^\circ)\), the stone hits the ball after time \(T_1\), and when \((\theta_0, \theta_1) = (60^\circ, 30^\circ)\), it hits the ball after time \(T_2\). In such a case, \(\left(\frac{T_1}{T_2}\right)^2\) is \_\_\_\_.

Hint:

For projectile motion problems involving relative velocities, use trigonometric identities to simplify time equations.

Answer 1

The time taken to hit is: \[ t = \frac{S_{\text{rel}}}{v_{\text{rel}}}. \] (I) For \((\theta_0, \theta_1) = (45^\circ, 45^\circ)\): \[ t = \frac{L}{v_0 \cos(45^\circ) + v_0 \cos(45^\circ)} = \frac{L}{2v_0 \cos(45^\circ)} = T_1. \] (II) For \((\theta_0, \theta_1) = (60^\circ, 30^\circ)\): \[ t = \frac{L}{v_0 \cos(60^\circ) + v_0 \cos(30^\circ)} = \frac{L}{2v_0}. \] Solving (I) and (II): \[ \left(\frac{T_1}{T_2}\right)^2 = 2. \]