An object A is kept fixed at the point x = 3 m and y = 1.25 m on a plank P raised above the ground.

Q: Wait, why is the vertical velocity negative at the time of collision?

Because the problem explicitly states the stone hits the object during its *downward* motion. In our standard coordinate system, upwards is positive and downwards is negative.

Q: Why did we multiply the velocity equation by $t$?

It's a beautiful algebraic trick! We had expressions for $u_x t$ and $u_y t$ from the displacement equations. By multiplying $u_x + u_y = 10t$ by $t$, we instantly created those exact terms, allowing for a seamless substitution without solving for $u_x$ and $u_y$ individually first.

Q: Can't we just use the equation of trajectory here?

You could, but it gets incredibly messy because the target is moving! The equation of trajectory $y = x \tan\theta - \dots$ assumes a static frame. Since the $x$-coordinate of the target is a function of time, parameterizing both $x$ and $y$ with respect to time $t$ is much cleaner.

Q: Why is the initial $x$-position of the plank $3$ m and not $0$?

The problem states the object is kept fixed at $x = 3$ m on the plank. The origin $(0,0)$ is where the stone is projected from. Always establish a single, absolute coordinate system for all objects in the problem.

Q: What if the stone hit the object during its upward motion?

If it hit during the upward motion at $45^\circ$, the vertical velocity $v_y$ would be positive. The condition would then be $v_y = v_x$, leading to $u_y - 10t = u_x$. The rest of the algebraic process would be identical, just with a different sign!

Question

An object $A$ is kept fixed at the point $x = 3	ext{ m}$ and $y = 1.25	ext{ m}$ on a plank $P$ raised above the ground. At time $t = 0$, the plank starts moving along the $+x$-direction with an acceleration $1.5	ext{ m/s}^2$. At the same instant, a stone is projected from the origin with a velocity $\mathbf{u}$ as shown. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of $45^\circ$ to the horizontal. All the motions are in $x$-$y$ plane. Find $\mathbf{u}$ and the time after which the stone hits the object. (Take $g = 10	ext{ m/s}^2$).

Sigma Percentile · Accepted Answer

The beauty of kinematics lies in its ability to predict the future. In this thrilling problem, we are not just analyzing a static scene; we are orchestrating a perfect, mid-air collision between a flying stone and an accelerating target. Imagine the precision required! A plank is speeding away, and we must launch a stone from the origin with the exact velocity needed to strike a specific object on that plank at a very specific angle.

This is a classic JEE-level constraint problem. It tests your ability to break down complex 2D motion into independent 1D components and then weave them back together using the thread of time. Let's dive into the physics and unravel this beautiful puzzle step by step.

### Analyzing the Setup

First, let's establish our coordinate system. The stone is projected from the origin, so its initial position is $(0,0)$. Let its initial velocity be $\mathbf{u} = u_x \hat{i} + u_y \hat{j}$.

The target, object $A$, is not at the origin. It is sitting on a plank at an initial position of $x = 3$ m and $y = 1.25$ m. But it's not staying there! At the exact moment the stone is launched ($t = 0$), the plank begins to accelerate in the positive $x$-direction with an acceleration of $a = 1.5$ m/s$^2$.

Our goal is to find the initial velocity $\mathbf{u}$ and the time $t$ when the stone perfectly strikes object $A$. For a collision to occur, the stone and the object must occupy the exact same spatial coordinates $(x, y)$ at the exact same time $t$.

### The Vertical Journey

Let's isolate the vertical motion. This is the easiest part because the plank is only moving horizontally. The object $A$ remains at a constant height of $y = 1.25$ m throughout its journey.

Therefore, for the stone to hit the object, its vertical displacement at the time of collision $t$ must be exactly $1.25$ m. We can use the second equation of motion for constant acceleration:

$$S_y = u_y t + \frac{1}{2} a_y t^2$$

Here, the vertical acceleration is solely due to gravity, acting downwards, so $a_y = -g = -10$ m/s$^2$. Substituting our known values, we get:

$$1.25 = u_y t - \frac{1}{2} (10) t^2$$

Simplifying this, we obtain our first master equation:

$$u_y t = 1.25 + 5 t^2$$

We will call this **Equation 1**. Notice how we isolated the term $u_y t$. This is a strategic move that will pay off beautifully later.

### The Horizontal Chase

Now, let's look at the horizontal motion. This is a chase! The stone is moving horizontally with a constant velocity $u_x$, while the object $A$ is accelerating away from it.

The horizontal position of the stone at time $t$ is simply:

$$x_{	ext{stone}} = u_x t$$

The horizontal position of object $A$ is a bit more complex. It starts at $x_0 = 3$ m and accelerates from rest relative to the plank's initial state. Using the kinematic equation for position:

$$x_A = x_0 + u_{Ax} t + \frac{1}{2} a_x t^2$$

Substituting the given values ($x_0 = 3$, $u_{Ax} = 0$, $a_x = 1.5$):

$$x_A = 3 + \frac{1}{2} (1.5) t^2 = 3 + 0.75 t^2$$

For the collision to happen, $x_{	ext{stone}}$ must equal $x_A$. Equating them gives us our second master equation:

$$u_x t = 3 + 0.75 t^2$$

We will call this **Equation 2**. Again, we have isolated the term $u_x t$.

### The Master Clue

We have two equations but three unknowns ($u_x$, $u_y$, and $t$). We need one more piece of information, and the problem provides a brilliant one: *A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of $45^\circ$ to the horizontal.*

This is the golden key. The angle of the velocity vector $\mathbf{v}$ at any instant is given by $	an	heta = \frac{v_y}{v_x}$. Since the stone is moving downwards at $45^\circ$, the angle is $-45^\circ$.

$$	an(-45^\circ) = \frac{v_y}{v_x} = -1$$

This beautifully simplifies to:

$$v_y = -v_x$$

Now, let's express the velocity components $v_x$ and $v_y$ at time $t$ in terms of our initial variables. The horizontal velocity remains constant, so $v_x = u_x$. The vertical velocity changes due to gravity, so $v_y = u_y - gt = u_y - 10t$.

Substituting these into our condition $v_y = -v_x$:

$$u_y - 10t = -u_x$$

Rearranging this gives us our third master equation:

$$u_x + u_y = 10t$$

### The Algebraic Symphony

Now comes the moment of algebraic elegance. We have:
1. $u_y t = 1.25 + 5 t^2$
2. $u_x t = 3 + 0.75 t^2$
3. $u_x + u_y = 10t$

We *could* solve Equation 3 for $u_x$ and substitute it into the others, but that gets messy. Instead, look at Equations 1 and 2. They contain $u_x t$ and $u_y t$. How can we transform Equation 3 to match?

By simply multiplying the entire Equation 3 by $t$!

$$(u_x + u_y) t = 10t \cdot t$$
$$u_x t + u_y t = 10 t^2$$

This is a stroke of genius. Now, we can directly substitute the expressions from Equation 1 and Equation 2 into this new equation:

$$(3 + 0.75 t^2) + (1.25 + 5 t^2) = 10 t^2$$

### Final Calculation

We have successfully eliminated $u_x$ and $u_y$, leaving a single equation with only one variable: time $t$. Let's combine the like terms:

$$4.25 + 5.75 t^2 = 10 t^2$$

Subtracting $5.75 t^2$ from both sides:

$$4.25 = 4.25 t^2$$

Dividing by $4.25$:

$$t^2 = 1$$

Since time must be positive, we find the time of collision:

$$t = 1 	ext{ s}$$

The hard part is over! Now we just reap the rewards. We substitute $t = 1$ back into Equations 1 and 2 to find the initial velocity components.

From Equation 1:
$$u_y (1) = 1.25 + 5(1)^2 = 6.25 	ext{ m/s}$$

From Equation 2:
$$u_x (1) = 3 + 0.75(1)^2 = 3.75 	ext{ m/s}$$

Therefore, the initial velocity vector of the stone must be:

$$\mathbf{u} = (3.75 \hat{i} + 6.25 \hat{j}) 	ext{ m/s}$$

And there we have it! By carefully breaking the motion into independent axes, translating physical constraints into mathematical equations, and using a clever algebraic trick, we have solved a complex 2D kinematics problem. This is the true power of physics!

Visualized Solution

Visualizing the Setup

Vertical Displacement of Stone

Formulating Vertical Equation

Simplifying Vertical Equation

Horizontal Displacement Condition

Formulating Horizontal Equation

Simplifying Horizontal Equation

Velocity Condition at Collision

Expressing Velocity Components

Applying the $4 5^{\circ}$ Condition

Multiplying by Time $t$

Substituting Displacements

Solving for Time $t$

Calculating Initial Velocity

Final Answer

The Sigma Insight: Motion in a Plane

Analyzing the Setup

The Vertical Journey

The Horizontal Chase

The Master Clue

The Algebraic Symphony

Final Calculation

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