A ball is thrown from a point with a speed v_0 at an angle of projection theta. From the same point

Q: Why don't we consider the vertical motion of the ball?

The vertical motion determines the maximum height and the total time of flight. However, whether the person catches the ball or not depends entirely on whether they can reach the landing spot at the same time. By matching horizontal velocities, the person ensures they are always exactly under the ball, guaranteeing a catch regardless of the time of flight.

Q: What if the person's speed was given as $2v_0$?

If the person's speed was $2v_0$, equating the velocities would give $\cos\theta = 2$. Since the cosine function can never exceed 1, this equation has no real solution. Physically, this means the person is running too fast and will outrun the ball.

Q: Does the mass of the ball matter in this problem?

No, the mass of the ball does not matter. In ideal projectile motion (ignoring air resistance), the trajectory and kinematics are completely independent of the object's mass.

Q: Why did $v_0$ cancel out in the equation?

$v_0$ canceled out because the runner's speed was given as a proportional fraction of the ball's initial speed ($\frac{v_0}{2}$). This means the required angle depends only on this ratio, not on how hard the ball was actually thrown.

Q: What if the person started running a bit later instead of at the same instant?

If the person started later, their velocities wouldn't need to match exactly at all times. Instead, you would need to equate their total horizontal displacements at the specific time the ball lands (Time of Flight). The math would become significantly more complex!

Question

A ball is thrown from a point with a speed $v_0$ at an angle of projection $	heta$. From the same point and at the same instant, a person starts running with a constant speed $\frac{v_0}{2}$ to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?

Options:

(A) Yes, $60^\circ$

(B) Yes, $30^\circ$

(C) No

(D) Yes, $45^\circ$

Sigma Percentile · Accepted Answer

### The Physics of a Catch

Imagine you are standing on a field. You throw a ball in the air with an initial speed $v_0$ at an angle $	heta$. At that exact same moment, you start sprinting forward to catch it. Your running speed is exactly half of the ball's initial speed, meaning $v_{	ext{person}} = \frac{v_0}{2}$.

Now, think about it... what is the physical condition required for you to catch the ball? Well, you and the ball must cover the exact same horizontal distance in the same amount of time. Since you both start at the same position $x=0$ at time $t=0$, your horizontal positions must match at all times. This implies that your horizontal velocity must perfectly match the ball's horizontal velocity.

### The Mathematical Setup

Let's write this down mathematically. When a projectile is launched, its velocity can be resolved into two independent components. The vertical component $v_0 \sin	heta$ dictates how high the ball goes and how long it stays in the air. The horizontal component, however, dictates how fast it moves across the ground.

The horizontal component of the ball's velocity is given by:
$$v_{x,	ext{ball}} = v_0 \cos	heta$$

And your running speed is given as:
$$v_{	ext{person}} = \frac{v_0}{2}$$

For you to always remain exactly underneath the ball (and eventually catch it), these two velocities must be equal. Let's equate them:
$$v_0 \cos	heta = \frac{v_0}{2}$$

### The Elegant Solution

Look closely at this equation. The $v_0$ term is present on both sides. Because $v_0$ is non-zero, we can safely divide both sides by $v_0$, and it beautifully cancels out. We are left with a very simple trigonometric equation:
$$\cos	heta = \frac{1}{2}$$

And we all know from our basic trigonometry that the cosine of $60^\circ$ is $\frac{1}{2}$. Therefore, the angle of projection must be exactly:
**$$	heta = 60^\circ$$**

So, yes, it is absolutely possible for the person to catch the ball, provided they throw it at a steep $60^\circ$ angle. This ensures the ball doesn't travel horizontally faster than they can run!

Select Answer:

Visualized Solution

$Visualizing the Scenario$

$Condition for Catching$

$Equating Velocities$

$Solving for θ$

$Final Answer$

$The Way Forward$

The Sigma Insight: Projectile Motion

The Physics of a Catch

The Mathematical Setup

The Elegant Solution

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Scenario

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Catching

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Velocities

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for θ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward