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

To solve the problem, we need to determine the time at which the stone hits the ball, given their initial velocities and launch angles.

1. Equations of Motion:
Let the position of the ball at time $t$ be $(x_b, y_b)$, and the position of the stone be $(x_s, y_s)$. The equations for the motion of the ball and the stone are:

Ball: $x_b = v_0 \cos\theta_0 t$ and $y_b = v_0 \sin\theta_0 t - \frac{1}{2}gt^2$

Stone: $x_s = L - v_s \cos\theta_1 t$ and $y_s = v_s \sin\theta_1 t - \frac{1}{2}gt^2$

2. Conditions for the Stone to Hit the Ball:
For the stone to hit the ball, their positions must be equal at some time $t$, i.e., $x_b = x_s$ and $y_b = y_s$ at that time.

3. Solving for $t$:
From the $x$-position equation:

$$ v_0 \cos\theta_0 t = L - v_s \cos\theta_1 t \Rightarrow v_0 \cos\theta_0 t + v_s \cos\theta_1 t = L $$

From the $y$-position equation:

$$ v_0 \sin\theta_0 t - \frac{1}{2}gt^2 = v_s \sin\theta_1 t - \frac{1}{2}gt^2 \Rightarrow v_0 \sin\theta_0 t = v_s \sin\theta_1 t $$

So we have:

$$ v_s = \frac{v_0 \sin\theta_0}{\sin\theta_1} $$

4. Substituting into the $x$-equation:
Substituting $v_s = \frac{v_0 \sin\theta_0}{\sin\theta_1}$ into the $x$-equation:

$$ v_0 \cos\theta_0 t + \frac{v_0 \sin\theta_0}{\sin\theta_1} \cos\theta_1 t = L $$

Factor out $v_0$:

$$ v_0 (\cos\theta_0 + \sin\theta_0 \cot\theta_1)t = L $$

Solving for $t$, we get:

$$ t = \frac{L}{v_0 (\cos\theta_0 + \sin\theta_0 \cot\theta_1)} $$

5. Case 1: $\theta_0 = 45^\circ$, $\theta_1 = 45^\circ$:
For $\theta_0 = 45^\circ$ and $\theta_1 = 45^\circ$, we have:

$$ t = T_1 = \frac{L}{v_0 \left(\cos 45^\circ + \sin 45^\circ \cot 45^\circ \right)} = \frac{L}{v_0 \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} (1)\right)} = \frac{L}{v_0 \frac{2}{\sqrt{2}}} = \frac{L}{v_0 \sqrt{2}} $$

6. Case 2: $\theta_0 = 60^\circ$, $\theta_1 = 30^\circ$:
For $\theta_0 = 60^\circ$ and $\theta_1 = 30^\circ$, we have:

$$ t = T_2 = \frac{L}{v_0 \left(\cos 60^\circ + \sin 60^\circ \cot 30^\circ\right)} = \frac{L}{v_0 \left(\frac{1}{2} + \frac{\sqrt{3}}{2} (\sqrt{3})\right)} = \frac{L}{v_0 \left(\frac{1}{2} + \frac{3}{2}\right)} = \frac{L}{v_0 (2)} $$

7. Ratio of Times:
Now, we compute the ratio of the times:

$$ \left(\frac{T_1}{T_2}\right)^2 = \left(\frac{\frac{L}{v_0 \sqrt{2}}}{\frac{L}{2v_0}}\right)^2 = \left(\frac{2}{\sqrt{2}}\right)^2 = (\sqrt{2})^2 = 2 $$

8. Final Answer:
The final answer is $\boxed{2}$.