Let's break down the problem: ### **Question:** A ball is thrown at an angle of 30° 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. #### **Given:** - Height of building, \( h = 20 \) m - Distance from building to P, \( R = 100 \) m - Angle of projection, \( \theta = 30^\circ \) Let's assume the ball is thrown with initial speed \( u \). --- ## **Step 1: Horizontal Motion** \[ R = u \cos\theta \cdot t \] \[ 100 = u \cos 30^\circ \cdot t \] \[ 100 = u \left(\frac{\sqrt{3}}{2}\right) t \] \[ u t = \frac{200}{\sqrt{3}} \] --- ## **Step 2: Vertical Motion** Equation of motion: \[ y = u \sin\theta \cdot t - \frac{1}{2}gt^2 \] Since the ball starts from height 20 m and lands on the ground (\( y = -20 \)): \[ -20 = u \sin 30^\circ \cdot t - \frac{1}{2}gt^2 \] \[ -20 = u \left(\frac{1}{2}\right)t - \frac{1}{2} \cdot 9.8 t^2 \] \[ -20 = \frac{u t}{2} - 4.9 t^2 \] From Step 1: \( u t = \frac{200}{\sqrt{3}} \) Plug into above: \[ -20 = \frac{1}{2} \cdot \frac{200}{\sqrt{3}} - 4.9 t^2 \] \[ -20 = \frac{100}{\sqrt{3}} - 4.9 t^2 \] \[ -20 - \frac{100}{\sqrt{3}} = -4.9 t^2 \] \[ 4.9 t^2 = \frac{100}{\sqrt{3}} + 20 \] \[ t^2 = \frac{\frac{100}{\sqrt{3}} + 20}{4.9} \] Calculate numerically: \[ \sqrt{3} \approx 1.732 \] \[ \frac{100}{1.732} \approx 57.74 \] \[ t^2 = \frac{57.74 + 20}{4.9} = \frac{77.74}{4.9} \approx 15.87 \] \[ t \approx \sqrt{15.87} \approx 3.99~\text{sec} \] --- ## **Step 3: Check the Options** Options visible: - 3.98 sec - 5.98 sec - 2 sec - 25 sec ### **Closest Answer:** **3.98 sec** --- ## **Final Answer:** **Correct option:** **3.98 sec**