Two towers AB and CD are situated a distance d apart as shown in figure. AB is 20 m high and CD is 3

Q: Why is the relative acceleration zero?

Both objects are projectiles moving solely under the influence of gravity. Therefore, the acceleration of object A is $-g \hat{j}$ and the acceleration of object C is also $-g \hat{j}$. When you calculate the relative acceleration ($\mathbf{a}_C - \mathbf{a}_A$), the gravity terms perfectly cancel out, leaving zero. This means in the frame of reference of one object, the other appears to move at a constant velocity in a straight line.

Q: Why must the relative velocity be parallel to the relative position?

Imagine you are sitting on object A. Because relative acceleration is zero, object C appears to be moving towards you in a straight line at a constant speed. For object C to actually hit you, its velocity vector (the direction it's moving) must point exactly along the line connecting you and object C (the relative position vector). If it points anywhere else, it will miss you.

Q: Why did we take the x-component of velocity for C as negative?

We chose a standard Cartesian coordinate system where the origin is at the base of tower A, with the positive x-axis pointing to the right (towards tower C). Since object C is thrown from tower C *towards* tower A, it is moving to the left, which is the negative x-direction. Therefore, its horizontal velocity component must be negative.

Q: What happens to the y-momentum during the collision?

The y-momentum is also conserved *during the infinitesimally short time of the collision*. However, unlike the x-direction where there are no external forces, gravity is constantly acting in the y-direction. So while we could calculate the exact vertical velocity immediately after the collision using momentum conservation, we don't need to! The problem only asks for the horizontal position where they land, and since the horizontal velocity becomes zero, the vertical velocity doesn't affect the landing spot's x-coordinate.

Q: Why doesn't the combined mass follow a parabolic path after collision?

A parabolic path requires a horizontal component of velocity. When we calculated the total horizontal momentum of the system just before the collision, it turned out to be exactly zero ($10m$ from A and $-10m$ from C). Because they stick together (inelastic collision), the combined mass must also have zero horizontal momentum, and thus zero horizontal velocity. With only gravity acting on it and no sideways motion, it falls straight down.

Question

Two towers $AB$ and $CD$ are situated a distance $d$ apart as shown in figure. $AB$ is $20	ext{ m}$ high and $CD$ is $30	ext{ m}$ high from the ground. An object of mass $m$ is thrown from the top of $AB$ horizontally with a velocity of $10	ext{ m/s}$ towards $CD$. Simultaneously, another object of mass $2m$ is thrown from the top of $CD$ at an angle of $60^\circ$ to the horizontal towards $AB$ with the same magnitude of initial velocity as that of the first object. The two objects move in the same vertical plane, collide in mid-air and stick to each other. (1994)
(a) Calculate the distance $d$ between the towers.
(b) Find the position where the objects hit the ground.

Sigma Percentile · Accepted Answer

### The Dance of Relative Motion

Imagine standing on a vast, flat plain with two towering structures rising before you. Tower $AB$ stands at $20	ext{ m}$, and tower $CD$ looms higher at $30	ext{ m}$. From the top of these towers, two objects are launched simultaneously. Object $A$ is thrown purely horizontally towards $CD$ at $10	ext{ m/s}$. Object $C$ is hurled downwards at an angle of $60^\circ$ towards $AB$, also with an initial speed of $10	ext{ m/s}$. They are destined to collide in mid-air. Our mission? To find the exact distance $d$ between the towers and the precise spot where their combined wreckage hits the ground.

At first glance, tracking two parabolic trajectories simultaneously seems like a mathematical nightmare. But physics offers us a beautiful shortcut: **Relative Motion**.

### The Magic of the Relative Frame

Once the objects leave the hands of the throwers, they are both in free fall. The only force acting on them is gravity, meaning they both experience a downward acceleration of $g$.

If we jump into the reference frame of object $A$, what do we see? Since both objects are accelerating downwards at exactly the same rate, their relative acceleration is zero:

$$ \mathbf{a}_{CA} = \mathbf{a}_C - \mathbf{a}_A = (-g \hat{j}) - (-g \hat{j}) = 0 $$

This is a profound realization. In the relative frame, the complex parabolic dance vanishes. From $A$'s perspective, object $C$ is moving in a perfectly straight line at a constant velocity!

### Setting Up the Vectors

Let's establish a coordinate system with the base of tower $AB$ (point $B$) as our origin $(0,0)$.

The velocity of object $A$ is purely horizontal:
$$ \mathbf{v}_A = 10 \hat{i} $$

Object $C$ is thrown downwards at $60^\circ$. We must resolve this into its $x$ and $y$ components. Since it's moving leftwards (towards $A$) and downwards, both components will be negative:
$$ \mathbf{v}_C = -10 \cos 60^\circ \hat{i} - 10 \sin 60^\circ \hat{j} $$
$$ \mathbf{v}_C = -5 \hat{i} - 5\sqrt{3} \hat{j} $$

Now, we calculate the relative velocity of $C$ with respect to $A$:
$$ \mathbf{v}_{CA} = \mathbf{v}_C - \mathbf{v}_A = (-5 \hat{i} - 5\sqrt{3} \hat{j}) - (10 \hat{i}) $$
$$ \mathbf{v}_{CA} = -15 \hat{i} - 5\sqrt{3} \hat{j} $$

Next, we need their relative position. Object $A$ starts at $(0, 20)$ and object $C$ starts at $(d, 30)$. The relative position vector is:
$$ \mathbf{r}_{CA} = \mathbf{r}_C - \mathbf{r}_A = (d - 0)\hat{i} + (30 - 20)\hat{j} $$
$$ \mathbf{r}_{CA} = d \hat{i} + 10 \hat{j} $$

### The Collision Condition

For a collision to occur, object $C$ must travel directly towards object $A$ in the relative frame. Mathematically, this means the relative velocity vector $\mathbf{v}_{CA}$ must be perfectly parallel to the relative position vector $\mathbf{r}_{CA}$.

This implies that the time taken to cover the relative horizontal distance must equal the time taken to cover the relative vertical distance:
$$ t = \frac{x_{rel}}{|v_{x,rel}|} = \frac{y_{rel}}{|v_{y,rel}|} $$

Let's use the vertical components to find the time of collision $t$:
$$ t = \frac{10}{5\sqrt{3}} = \frac{2}{\sqrt{3}} 	ext{ s} $$

Now, we can use this time to find the unknown distance $d$ between the towers:
$$ d = |v_{x,rel}| 	imes t = 15 	imes \frac{2}{\sqrt{3}} = \frac{30}{\sqrt{3}} $$
$$ d = 10\sqrt{3} \approx 17.32 	ext{ m} $$

### The Inelastic Embrace

We now know when and where the collision happens. The $x$-coordinate of the collision point $P$ is simply the horizontal distance traveled by object $A$ in time $t$:
$$ x_P = v_{Ax} 	imes t = 10 	imes \frac{2}{\sqrt{3}} = \frac{20}{\sqrt{3}} \approx 11.55 	ext{ m} $$

The problem states that upon colliding, the two objects stick together. This is a classic perfectly inelastic collision. To find out what happens next, we must invoke the **Conservation of Linear Momentum**.

Let's look at the horizontal ($x$) direction. The mass of object $A$ is $m$, and the mass of object $C$ is $2m$. The initial horizontal momentum of the system just before the collision is:
$$ p_{ix} = m(v_{Ax}) + 2m(v_{Cx}) $$
$$ p_{ix} = m(10) + 2m(-5) = 10m - 10m = 0 $$

This is a stunning result! The horizontal momentum of the system is exactly zero. Because momentum is conserved, the final horizontal momentum of the combined mass $(3m)$ must also be zero.

$$ (3m)v_{fx} = 0 \implies v_{fx} = 0 $$

With absolutely no horizontal velocity remaining, the tangled wreckage of the two objects will simply drop straight down from the point of collision, plummeting vertically to the ground.

Therefore, the final resting place of the objects is exactly directly below the collision point $P$. The distance from the base of tower $AB$ (point $B$) is simply the $x$-coordinate we calculated earlier:

**Distance from B = $11.55	ext{ m}$**.

By masterfully blending relative kinematics with momentum conservation, a seemingly chaotic two-body problem collapses into an elegant and satisfying solution.

Visualized Solution

$Visualizing the Setup$

$a_{r e l} = 0$

$v_{A}$

$v_{C}$

$Components of v_{C}$

$v_{C A}$

$r_{C A}$

$v_{C A} ∥ r_{C A}$

$t = \frac{y _{r e l}}{v _{y, r e l}}$

$d = ∣ v_{x, r e l} ∣ \times t$

$x_{P} = v_{A x} \times t$

$p_{i x} = p_{f x}$

$v_{f x} = 0$

$Final Position$

The Sigma Insight: Relative Velocity

The Dance of Relative Motion

The Magic of the Relative Frame

Setting Up the Vectors

The Collision Condition

The Inelastic Embrace

Similar Questions

List I describes four systems, each with two particles A and B in relative motion as shown in figure. List II gives possible magnitudes of their relative velocities (in $ms^{- 1}$ ) at time $t = \frac{π}{3} s$ .

List-I

List-II

Two material particles A and B are moving in free space. How their position coordinates x, y and z vary with time t is shown in the following graphs. Determine at what instant of time the particles are closest to each other and the closest separation.

Comprehension Passage

When and for how long can the particle C fulfil the given condition?

What is speed of the particle C at the instant $t = 5 s$ ?

What is modulus of acceleration of the particle C at the instant $t = 5 s$ ?

At the instant, when the line joining locations of A and B is perpendicular to the line joining locations of B and C, what are the magnitudes of velocities of C relative to A and B respectively?

Visualized Solution

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Components of vC​

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vCA​∥rCA​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=vy,rel​yrel​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d=∣vx,rel​∣×t

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xP​=vAx​×t

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

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

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