An object is launched from a cliff `20 m` above the ground at an angle of `30^(@)` above the horizontal with an initial speed of `30m//s`.How far does the object travel before landing on the ground?(in metre)

Options

A. `20`
B. `20sqrt3`
C. `60`
D. `60sqrt3`

Explanation

To solve the problem, we will break it down into steps to find out how far the object travels before landing on the ground.

### Step 1: Identify the given data
- Height of the cliff (h) = 20 m
- Angle of projection (θ) = 30 degrees
- Initial speed (u) = 30 m/s
- Acceleration due to gravity (g) = 9.81 m/s² (approximately 10 m/s² for simplicity)

### Step 2: Resolve the initial velocity into components
The initial velocity can be resolved into horizontal (u_x) and vertical (u_y) components using trigonometric functions:
- \( u_x = u \cdot \cos(θ) = 30 \cdot \cos(30^\circ) = 30 \cdot \frac{\sqrt{3}}{2} = 15\sqrt{3} \, \text{m/s} \)
- \( u_y = u \cdot \sin(θ) = 30 \cdot \sin(30^\circ) = 30 \cdot \frac{1}{2} = 15 \, \text{m/s} \)

### Step 3: Determine the time of flight (t)
We will use the vertical motion equation to find the time of flight. The equation for vertical motion is:
\[ h = u_y \cdot t - \frac{1}{2} g t^2 \]
Substituting the values we have:
\[ 20 = 15t - \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ 20 = 15t - 5t^2 \]
Rearranging gives:
\[ 5t^2 - 15t + 20 = 0 \]
Dividing the entire equation by 5:
\[ t^2 - 3t + 4 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -3, c = 4 \):
\[ t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \]
\[ t = \frac{3 \pm \sqrt{9 - 16}}{2} \]
\[ t = \frac{3 \pm \sqrt{-7}}{2} \]
Since the discriminant is negative, we made a mistake in rearranging the equation. Let's correct it.

Rearranging correctly:
\[ 5t^2 - 15t - 20 = 0 \]
Using the quadratic formula:
\[ t = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 5 \cdot (-20)}}{2 \cdot 5} \]
\[ t = \frac{15 \pm \sqrt{225 + 400}}{10} \]
\[ t = \frac{15 \pm \sqrt{625}}{10} \]
\[ t = \frac{15 \pm 25}{10} \]

Calculating the two possible values:
1. \( t = \frac{40}{10} = 4 \, \text{s} \)
2. \( t = \frac{-10}{10} = -1 \, \text{s} \) (not physically meaningful)

Thus, the time of flight is \( t = 4 \, \text{s} \).

### Step 5: Calculate the horizontal distance traveled (R)
The horizontal distance traveled can be calculated using:
\[ R = u_x \cdot t \]
Substituting the values:
\[ R = 15\sqrt{3} \cdot 4 \]
\[ R = 60\sqrt{3} \, \text{m} \]

### Final Answer
The object travels approximately \( 60\sqrt{3} \) meters before landing on the ground.