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 \(\theta_0\)\(\theta_0\) 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 \((180 − \theta_1 )\) from the +𝑥-direction with a suitable initial speed. For a fixed \(𝑣_0\) , when \((\theta_0 , \theta_1 ) = (45° , 45° ),\) the stone hits the ball after time \(𝑇_1\) , and when \((\theta_0 , \theta_1 ) = (60° , 30° )\), it hits the ball after time \(𝑇_2\) . In such a case,\( (\frac{𝑇1 }{𝑇2} )^2\) is ______.

Answer 1

Correct Answer: 2

Time Taken to Hit

(I) For \( (\theta_0, \theta_1) = (45^\circ, 45^\circ) \): 

\[ t = \frac{S_{\text{rel}}}{v_{\text{rel}}} \] \[ t = \frac{L}{v_0 \cos(45^\circ) + v_0 \cos(45^\circ)} \]

(II) For \( (\theta_0, \theta_1) = (60^\circ, 30^\circ) \):

\[ t = \frac{L}{2v_0 \cos(45^\circ)} = T_1 \] \[ t = \frac{L}{v_0 \cos(60^\circ) + v_0 \cos(30^\circ)} = \frac{L}{2v_0} \]

Solving (I) and (II):

Quick Tip:

\[ \frac{T_1}{T_2} = 2 \]