Let's break down the problem: **Question:** A ball is thrown at an angle of 30 degrees from a building of height 20 m. Calculate the time taken to reach point P on the ground, which is 100 m away from the building. ### Step-by-Step Solution: Let's denote: - Height of building (h) = 20 m - Horizontal distance (Range) to point P (R) = 100 m - Angle of projection (θ) = 30° - Acceleration due to gravity (g) = 9.8 m/s² Let the initial velocity be \( u \). #### 1. **Horizontal Motion:** \[ R = u \cos\theta \cdot t \] \[ 100 = u \cos 30^\circ \cdot t \] \[ 100 = u \left(\frac{\sqrt{3}}{2}\right) t \] \[ u t = \frac{200}{\sqrt{3}} \] #### 2. **Vertical Motion:** The displacement in vertical direction after time \( t \) (taking upward as positive and initial position as 20 m): \[ y = h + (u \sin\theta) t - \frac{1}{2}gt^2 \] At the point where the ball hits the ground, \( y = 0 \): \[ 0 = 20 + (u \sin 30^\circ) t - \frac{1}{2}gt^2 \] \[ 0 = 20 + \left(u \cdot \frac{1}{2}\right)t - 4.9 t^2 \] \[ 0 = 20 + \frac{u t}{2} - 4.9 t^2 \] From step 1, \( u t = \frac{200}{\sqrt{3}} \): Plug this into the vertical equation: \[ 0 = 20 + \frac{1}{2} \left(\frac{200}{\sqrt{3}}\right) - 4.9 t^2 \] \[ 0 = 20 + \frac{100}{\sqrt{3}} - 4.9 t^2 \] \[ 4.9 t^2 = 20 + \frac{100}{\sqrt{3}} \] Calculate \(\frac{100}{\sqrt{3}}\): \(\sqrt{3} \approx 1.732\) \(\frac{100}{1.732} \approx 57.74\) So, \[ 4.9 t^2 = 20 + 57.74 = 77.74 \] \[ t^2 = \frac{77.74}{4.9} \approx 15.87 \] \[ t = \sqrt{15.87} \approx 3.99 \text{ seconds} \] ### Closest Answer: None of the options match exactly, but the closest is **3.98 sec** (option 3). ### **Final Answer:** **3.98 sec**