Two towers $AB$ and $CD$ are situated a distance $d$ apart as shown in figure. $AB$ is $20\,\text{m}$ high and $CD$ is $30\,\text{m}$ high from the ground. An object of mass $m$ is thrown from the top of $AB$ horizontally with a velocity of $10\,\text{m/s}$ 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. (i) Calculate the distance '$d$' between the towers, and (ii) Find the position where the objects hit the ground.

Step 1: Set up the coordinate system.

Let the ground be the x-axis ($y = 0$) and the base of tower $AB$ be the origin ($x = 0$).

Step 2: Write the equations of motion for the first object.

Object 1 of mass $m$ is projected horizontally from the top of tower $AB$ at $x = 0, y = 20\text{ m}$.

$$x_1(t) = 10t$$ $$y_1(t) = 20 - 5t^2$$

Step 3: Write the equations of motion for the second object.

Object 2 of mass $2m$ is projected from the top of tower $CD$ at $x = d, y = 30\text{ m}$ at an angle of $60^\circ$ below the horizontal.

$$v_{2x} = -10\cos 60^\circ = -5$$ $$v_{2y} = -10\sin 60^\circ = -5\sqrt{3}$$ $$x_2(t) = d - 5t$$ $$y_2(t) = 30 - 5\sqrt{3}t - 5t^2$$

Step 4: Find the time of collision.

Equate the vertical coordinates $y_1(t) = y_2(t)$ at the instant of collision.

$$20 - 5t^2 = 30 - 5\sqrt{3}t - 5t^2$$ $$5\sqrt{3}t = 10$$ $$t = \frac{2}{\sqrt{3}}\text{ s}$$

Step 5: Calculate the distance between the towers.

Equate the horizontal coordinates $x_1(t) = x_2(t)$ at the collision time.

$$10t = d - 5t$$ $$d = 15t$$ $$d = 15\left(\frac{2}{\sqrt{3}}\right) = 10\sqrt{3}\text{ m} \approx 17.3\text{ m}$$

Step 6: Determine the velocities of both objects immediately before the collision.

Substitute $t = \frac{2}{\sqrt{3}}\text{ s}$ into the velocity components.

$$\vec{v}_1 = \left(10, -\frac{20}{\sqrt{3}}\right)$$ $$\vec{v}_2 = \left(-5, -5\sqrt{3} - \frac{20}{\sqrt{3}}\right)$$

Step 7: Apply conservation of momentum for the collision.

The two objects stick together to form a combined mass of $3m$ with velocity $\vec{V} = (V_x, V_y)$.

$$m\vec{v}_1 + 2m\vec{v}_2 = 3m\vec{V}$$ $$V_x = \frac{10 + 2(-5)}{3} = 0$$ $$V_y = \frac{-\frac{20}{\sqrt{3}} + 2\left(-5\sqrt{3} - \frac{20}{\sqrt{3}}\right)}{3} = -10\sqrt{3}$$

Step 8: Find the landing position of the combined object.

Since the horizontal velocity component $V_x = 0$, the combined object falls vertically downward from the collision point.

$$x_{\text{collision}} = 10t = 10\left(\frac{2}{\sqrt{3}}\right) = \frac{20}{\sqrt{3}}\text{ m} \approx 11.5\text{ m}$$

Answer: (i) The distance between the towers is $d = 10\sqrt{3}\text{ m} \approx 17.3\text{ m}$. (ii) The objects hit the ground at a distance of $\frac{20}{\sqrt{3}}\text{ m} \approx 11.5\text{ m}$ from the base of tower $AB$.