A bomb is fired from a cannon with a velocity of `1000 ms^(-1)` making an angle of `30^(@)` with the horizontal `(g = 9.8 m//s^(2)).`
At what distance from the cannon the bomb will hit the ground?

Explanation

To find the distance from the cannon where the bomb will hit the ground, we will follow these steps:

### Step 1: Break down the initial velocity into components
The initial velocity \( v_0 = 1000 \, \text{m/s} \) is at an angle \( \theta = 30^\circ \) with the horizontal. We can find the horizontal and vertical components of the initial velocity using trigonometric functions.

- Horizontal component \( v_{0x} = v_0 \cos(\theta) = 1000 \cos(30^\circ) \)
- Vertical component \( v_{0y} = v_0 \sin(\theta) = 1000 \sin(30^\circ) \)

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