Let's break down the question step by step: **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. Let's denote: - Initial velocity as \( u \) - Angle of projection, \( \theta = 30^\circ \) - Height of building, \( h = 20 \) m - Horizontal distance, \( R = 100 \) m - Acceleration due to gravity, \( g = 9.8 \) m/s² ### Horizontal motion: \[ R = u \cos \theta \cdot t \] \[ 100 = u \cos 30^\circ \cdot t \] \[ 100 = u \left(\frac{\sqrt{3}}{2}\right) t \] \[ u = \frac{200}{\sqrt{3} t} \] ### Vertical motion (using equation for displacement): \[ y = u \sin \theta \cdot t - \frac{1}{2} g t^2 \] Since the ball lands on the ground, and it starts from a height of 20 m: \[ -20 = u \sin 30^\circ \cdot t - \frac{1}{2} g t^2 \] \[ -20 = u \left(\frac{1}{2}\right)t - \frac{1}{2} \cdot 9.8 t^2 \] \[ -20 = \frac{u t}{2} - 4.9 t^2 \] Now substitute the value of \( u \) from earlier: \[ -20 = \frac{1}{2} \cdot \frac{200}{\sqrt{3} t} \cdot t - 4.9 t^2 \] \[ -20 = \frac{100}{\sqrt{3}} - 4.9 t^2 \] \[ -20 + 4.9 t^2 = \frac{100}{\sqrt{3}} \] \[ 4.9 t^2 = \frac{100}{\sqrt{3}} + 20 \] \[ t^2 = \frac{\frac{100}{\sqrt{3}} + 20}{4.9} \] \[ t^2 \approx \frac{57.74 + 20}{4.9} \] \[ t^2 \approx \frac{77.74}{4.9} \] \[ t^2 \approx 15.87 \] \[ t \approx 3.99 \text{ seconds} \] ### **So, the correct answer is:** **3.98 sec (rounded off to two decimals)** #### The correct option is: **3.98 sec**