A ball is thrown from the top of the Qutab Minar 200 ft. high with a velocity 80 ft. per second at an elevation of `30^(@)` above the horizon. The horizontal distance from the foot of the minar to. the point where it hits the ground is `(g= 32 ft.//"sec"^(2))`

Options

A. 200 ft.
B. ` 200 sqrt(3) ft.`
C. ` 300 sqrt(3) ft.`
D. ` 200 //sqrt(3) ft.`

Answer

To solve the problem of a ball thrown from the top of Qutab Minar, we will break the motion into horizontal and vertical components and use the equations of motion to find the horizontal distance traveled by the ball before it hits the ground.

### Step-by-step Solution:

1. **Identify the Given Data:**
- Height of Qutab Minar (h) = 200 ft
- Initial velocity (v) = 80 ft/s
- Angle of projection (θ) = 30°
- Acceleration due to gravity (g) = 32 ft/s²

2. **Calculate the Vertical and Horizontal Components of the Initial Velocity:**
- The vertical component (v_y) is given by:
\[ v_y = v \sin(θ) = 80 \sin(30°) = 80 \times \frac{1}{2} = 40 \text{ ft/s} \]
- The horizontal component (v_x) is given by:
\[ v_x = v \cos(θ) = 80 \cos(30°) = 80 \times \frac{\sqrt{3}}{2} = 40\sqrt{3} \text{ ft/s} \]

3. **Set Up the Vertical Motion Equation:**
- The vertical motion can be described by the equation:
\[ h = v_y t - \frac{1}{2} g t^2 \]
- Substituting the known values:
\[ -200 = 40t - \frac{1}{2} \times 32 t^2 \]
- Rearranging gives:
\[ 16t^2 - 40t - 200 = 0 \]

4. **Solve the Quadratic Equation:**
- Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- Here, \(a = 16\), \(b = -40\), and \(c = -200\).
- Calculate the discriminant:
\[ b^2 - 4ac = (-40)^2 - 4 \times 16 \times (-200) = 1600 + 12800 = 14400 \]
- Now substituting into the quadratic formula:
\[ t = \frac{40 \pm \sqrt{14400}}{2 \times 16} = \frac{40 \pm 120}{32} \]
- This gives two possible solutions for \(t\):
\[ t = \frac{160}{32} = 5 \text{ seconds} \quad \text{(valid, as time cannot be negative)} \]
\[ t = \frac{-80}{32} \quad \text{(not valid)} \]

5. **Calculate the Horizontal Distance:**
- The horizontal distance (R) can be calculated using:
\[ R = v_x \cdot t = 40\sqrt{3} \cdot 5 = 200\sqrt{3} \text{ ft} \]

### Final Answer:
The horizontal distance from the foot of the Qutab Minar to the point where the ball hits the ground is \(200\sqrt{3} \text{ ft}\).

---