A ball of mass $0.2\,\text{kg}$ rests on a vertical post of height $5\,\text{m}$. A bullet of mass $0.01\,\text{kg}$, traveling with a velocity $V\,\text{m/s}$ in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of $20\,\text{m}$ and the bullet at a distance of $100\,\text{m}$ from the foot of the post. The velocity $V$ of the bullet is

Step 1: Identify the given parameters.

Let the mass of the ball be $M$ and the mass of the bullet be $m$.

$$M = 0.2\,\text{kg}$$ $$m = 0.01\,\text{kg}$$ $$h = 5\,\text{m}$$ $$x_{\text{ball}} = 20\,\text{m}$$ $$x_{\text{bullet}} = 100\,\text{m}$$

Step 2: Calculate the time of flight.

Since both the ball and the bullet are projected horizontally from the same height, they have the same time of flight $t$.

$$h = \frac{1}{2}gt^2$$ $$t = \sqrt{\frac{2h}{g}}$$

Using $g = 10\,\text{m/s}^2$:

$$t = \sqrt{\frac{2(5)}{10}} = 1\,\text{s}$$

Step 3: Find the horizontal velocities after collision.

Calculate the horizontal velocity of the ball $v_{\text{ball}}$ and the bullet $v_{\text{bullet}}$ immediately after the collision.

$$v_{\text{ball}} = \frac{x_{\text{ball}}}{t} = \frac{20}{1} = 20\,\text{m/s}$$ $$v_{\text{bullet}} = \frac{x_{\text{bullet}}}{t} = \frac{100}{1} = 100\,\text{m/s}$$

Step 4: Apply conservation of linear momentum.

The total horizontal momentum before the collision equals the total horizontal momentum after the collision.

$$m V = M v_{\text{ball}} + m v_{\text{bullet}}$$

Substitute the known values into the momentum equation.

$$0.01 V = (0.2)(20) + (0.01)(100)$$ $$0.01 V = 4 + 1$$ $$0.01 V = 5$$ $$V = \frac{5}{0.01} = 500\,\text{m/s}$$

Answer: $500\,\text{m/s}$