A ball is thrown from the location (x0, y0) = (0, 0) of a horizontal playground with an initial spee

Q: Why do we ignore gravity in this problem?

The problem explicitly states the motion occurs on a 'horizontal playground'. This means we are looking at a 2D motion in the horizontal plane (like balls rolling on a flat table). Gravity acts perpendicular to this plane and therefore does not affect the x and y coordinates.

Q: Why must their y-velocities be exactly equal?

Both the ball and the stone start at $y=0$ at the exact same time. For them to collide, they must reach the same final y-coordinate at the same time $t$. Since $y = v_y t$, the only way for their y-displacements to be equal is if their y-velocities are equal.

Q: How did we get the relative velocity in the x-direction?

The ball is moving to the right (positive x-direction) and the stone is moving to the left (negative x-direction). Since they are moving towards each other, their relative speed of approach is the sum of the magnitudes of their x-velocity components.

Q: What if they were thrown at different times?

If there was a time delay, say the stone was thrown $\Delta t$ later, we would have to account for it in our displacement equations. The time for the stone would be $t - \Delta t$, making the algebra slightly more complex, but the core principle of equating coordinates would remain the same.

Q: Why is the angle for the stone given as $180^\circ - \theta_1$?

Angles are conventionally measured counter-clockwise from the positive x-axis. An angle of $180^\circ - \theta_1$ simply indicates that the stone is thrown from the right towards the left, making an interior angle of $\theta_1$ with the negative x-axis.

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 $	heta_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 $(180^\circ - 	heta_1)$ from the $+x$-direction with a suitable initial speed. For a fixed $v_0$, when $(	heta_0, 	heta_1) = (45^\circ, 45^\circ)$, the stone hits the ball after time $T_1$, and when $(	heta_0, 	heta_1) = (60^\circ, 30^\circ)$, it hits the ball after time $T_2$. In such a case, $(T_1/T_2)^2$ is _______________

Sigma Percentile · Accepted Answer

The beauty of kinematics lies in its ability to break down complex, multi-dimensional motion into simpler, independent one-dimensional components. This problem is a classic demonstration of that principle. We are presented with two objects—a ball and a stone—moving on a horizontal playground. The fact that the playground is horizontal is our first major clue: we are dealing with a purely two-dimensional motion in the $x-y$ plane, completely unbothered by the vertical pull of gravity.

### Analyzing the Setup

Imagine looking down at this playground from a bird's-eye view. The ball is launched from the origin $(0,0)$ with a velocity $v_0$ at an angle $	heta_0$. Simultaneously, a stone is launched from a point $(L,0)$ on the $x$-axis. It is thrown towards the ball, making an angle of $180^\circ - 	heta_1$ with the positive $x$-axis, which simply means it makes an angle $	heta_1$ with the negative $x$-axis.

For these two objects to collide, they must occupy the exact same spatial coordinates at the exact same time. This gives us two independent constraints: one for the $y$-axis and one for the $x$-axis.

### The Master Equations

Let's look at the $y$-axis first. Both objects start at $y=0$. For them to meet at some point $(x,y)$, their vertical displacements must be identical. Since there is no acceleration in this horizontal plane, their vertical velocities must be perfectly matched.

$$v_0 \sin 	heta_0 = v_s \sin 	heta_1$$

This equation is incredibly powerful. It tells us exactly how fast the stone must be thrown ($v_s$) to ensure it doesn't miss the ball vertically.

Now, let's consider the $x$-axis. The ball is moving to the right, and the stone is moving to the left. The total horizontal distance between them is initially $L$. To collide, the sum of the horizontal distances they cover must equal $L$. This is the classic concept of relative velocity of approach.

$$x_{	ext{ball}} + |x_{	ext{stone}}| = L$$

$$(v_0 \cos 	heta_0) t + (v_s \cos 	heta_1) t = L$$

Solving for the time of collision, $t$, we get our master equation:

$$t = \frac{L}{v_0 \cos 	heta_0 + v_s \cos 	heta_1}$$

### Case 1: The Symmetrical Throw

In the first scenario, both angles are $45^\circ$.

$$(	heta_0, 	heta_1) = (45^\circ, 45^\circ)$$

Plugging this into our $y$-axis constraint:

$$v_0 \sin 45^\circ = v_1 \sin 45^\circ \implies v_1 = v_0$$

The symmetry dictates that the stone must be thrown with the exact same speed as the ball. Now, we substitute this into our time equation to find $T_1$:

$$T_1 = \frac{L}{v_0 \cos 45^\circ + v_0 \cos 45^\circ}$$

$$T_1 = \frac{L}{v_0 \left(\frac{1}{\sqrt{2}}ight) + v_0 \left(\frac{1}{\sqrt{2}}ight)} = \frac{L}{\sqrt{2} v_0}$$

### Case 2: The Asymmetrical Throw

In the second scenario, the angles are different.

$$(	heta_0, 	heta_1) = (60^\circ, 30^\circ)$$

Again, we start with the $y$-axis constraint to find the required speed of the stone, $v_2$:

$$v_0 \sin 60^\circ = v_2 \sin 30^\circ$$

$$v_0 \left(\frac{\sqrt{3}}{2}ight) = v_2 \left(\frac{1}{2}ight) \implies v_2 = \sqrt{3} v_0$$

Because the stone is thrown at a shallower angle ($30^\circ$), it must be thrown much faster ($\sqrt{3}$ times faster) to keep up with the vertical pace of the ball. Now, we find the new collision time, $T_2$:

$$T_2 = \frac{L}{v_0 \cos 60^\circ + v_2 \cos 30^\circ}$$

$$T_2 = \frac{L}{v_0 \left(\frac{1}{2}ight) + \sqrt{3} v_0 \left(\frac{\sqrt{3}}{2}ight)}$$

$$T_2 = \frac{L}{\frac{v_0}{2} + \frac{3v_0}{2}} = \frac{L}{2v_0}$$

### Final Calculation

The problem asks for the square of the ratio of these two times. Let's divide $T_1$ by $T_2$:

$$\frac{T_1}{T_2} = \frac{\frac{L}{\sqrt{2} v_0}}{\frac{L}{2 v_0}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$

Finally, squaring this ratio gives us our elegant final answer:

**$$\left(\frac{T_1}{T_2}ight)^2 = 2$$**

This problem beautifully illustrates how setting up the right constraints—equating coordinates for a collision—turns a seemingly complex 2D kinematics problem into a straightforward algebraic exercise.

Enter Numerical Value:

Visualized Solution

$y_{ball} = y_{stone}$

$v_{0} sin θ_{0} = v_{s} sin θ_{1}$

$t = \frac{L}{v _{0} c o s θ _{0} + v _{s} c o s θ _{1}}$

$Case 1: θ_{0} = 4 5^{\circ}, θ_{1} = 4 5^{\circ}$

$T_{1} = \frac{L}{2 v _{0}}$

$Case 2: θ_{0} = 6 0^{\circ}, θ_{1} = 3 0^{\circ}$

$T_{2} = \frac{L}{2 v _{0}}$

$(T_{1} / T_{2})^{2} = 2$

\text{Reflections & Variations}

The Sigma Insight: Motion in a Plane

Analyzing the Setup

The Master Equations

Case 1

Case 2

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } yball​=ystone​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v0​sinθ0​=vs​sinθ1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=v0​cosθ0​+vs​cosθ1​L​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 1: θ0​=45∘,θ1​=45∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​=2​v0​L​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 2: θ0​=60∘,θ1​=30∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T2​=2v0​L​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (T1​/T2​)2=2