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 $\boxed{2}$.

Answer 2

To solve the problem, we need to compare the square of the ratio of times taken by a stone to hit a ball for two different sets of launch angles.

1. Understanding the Problem Setup:
- A ball is projected from the origin (0, 0) at speed $v_0$ and angle $\theta_0$.
- A stone is thrown from point $(L, 0)$ towards the ball with angle $(180^\circ - \theta_1)$.
- The collision happens in mid-air at time $T_1$ or $T_2$ depending on the angles.

2. Parametric Coordinates of Ball:
For the ball projected from origin at angle $\theta_0$ and speed $v_0$:
$x_b(t) = v_0 \cos \theta_0 \cdot t$
$y_b(t) = v_0 \sin \theta_0 \cdot t - \frac{1}{2}gt^2$

3. Parametric Coordinates of Stone:
Let the speed of stone be $v_s$, thrown at angle $(180^\circ - \theta_1)$:
$x_s(t) = L - v_s \cos \theta_1 \cdot t$
$y_s(t) = v_s \sin \theta_1 \cdot t - \frac{1}{2}gt^2$

4. Condition for Collision:
For the stone to hit the ball at time $t$, their x- and y-coordinates must be the same:
$v_0 \cos \theta_0 \cdot t = L - v_s \cos \theta_1 \cdot t$
$⇒ t (v_0 \cos \theta_0 + v_s \cos \theta_1) = L$
$⇒ t = \frac{L}{v_0 \cos \theta_0 + v_s \cos \theta_1}$

5. Matching y-coordinates:
$y_b(t) = y_s(t)$
Since vertical motion must also match at collision point, equating the vertical components:
$v_0 \sin \theta_0 = v_s \sin \theta_1$
$⇒ v_s = \frac{v_0 \sin \theta_0}{\sin \theta_1}$

6. Substituting $v_s$ into $t$:
$t = \frac{L}{v_0 \cos \theta_0 + \frac{v_0 \sin \theta_0}{\sin \theta_1} \cos \theta_1 }$
$= \frac{L}{v_0 \left[ \cos \theta_0 + \frac{\sin \theta_0 \cos \theta_1}{\sin \theta_1} \right]}$

7. Time Expressions:
Let’s define:
$T = \frac{L}{v_0 \cdot f(\theta_0, \theta_1)}$, where:
$f(\theta_0, \theta_1) = \cos \theta_0 + \frac{\sin \theta_0 \cos \theta_1}{\sin \theta_1}$

8. Compute $T_1$ and $T_2$:
- Case 1: $(\theta_0, \theta_1) = (45^\circ, 45^\circ)$
$f_1 = \cos 45^\circ + \frac{\sin 45^\circ \cos 45^\circ}{\sin 45^\circ} = \cos 45^\circ + \cos 45^\circ = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2}$

- Case 2: $(\theta_0, \theta_1) = (60^\circ, 30^\circ)$
$f_2 = \cos 60^\circ + \frac{\sin 60^\circ \cos 30^\circ}{\sin 30^\circ}$
$\cos 60^\circ = \frac{1}{2}$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\sin 30^\circ = \frac{1}{2}$
$f_2 = \frac{1}{2} + \frac{\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{1}{2} + \frac{3/4}{1/2} = \frac{1}{2} + \frac{3}{2} = 2$

9. Ratio of Times:
$T_1 \propto \frac{1}{f_1} = \frac{1}{\sqrt{2}}$
$T_2 \propto \frac{1}{f_2} = \frac{1}{2}$
$ \left( \frac{T_1}{T_2} \right)^2 = \left( \frac{1/\sqrt{2}}{1/2} \right)^2 = \left( \frac{2}{\sqrt{2}} \right)^2 = ( \sqrt{2} )^2 = 2 $

Final Answer:
$ \left( \frac{T_1}{T_2} \right)^2 = \boxed{2} $