Let's break down the problem: **Given:** - Initial height, \( h = 20 \) m - Angle of projection, \( \theta = 30^\circ \) - Horizontal distance, \( R = 100 \) m - We need to find the **time taken to reach point P** on the ground. We are not given the initial velocity (\( u \)), so let's use the equations of projectile motion: ### Step 1: Horizontal Motion \[ R = u \cos \theta \cdot t \] ### Step 2: Vertical Motion \[ y = h + (u \sin \theta)t - \frac{1}{2}gt^2 \] At point P, \( y = 0 \): \[ 0 = 20 + (u \sin 30^\circ)t - \frac{1}{2}gt^2 \] \[ 0 = 20 + (u \cdot 0.5)t - 4.9t^2 \] ### Step 3: Express \( u \) in terms of \( t \) using the horizontal equation: \[ u \cos 30^\circ = \frac{R}{t} \] \[ u \cdot \frac{\sqrt{3}}{2} = \frac{100}{t} \] \[ u = \frac{200}{\sqrt{3}t} \] ### Step 4: Substitute \( u \) in the vertical equation: \[ 0 = 20 + \left(\frac{200}{\sqrt{3}t} \cdot 0.5\right)t - 4.9t^2 \] \[ 0 = 20 + \frac{100}{\sqrt{3}} - 4.9t^2 \] \[ 4.9t^2 = 20 + \frac{100}{\sqrt{3}} \] \[ t^2 = \frac{20 + \frac{100}{\sqrt{3}}}{4.9} \] \[ \sqrt{3} \approx 1.732 \] \[ \frac{100}{1.732} \approx 57.74 \] \[ 20 + 57.74 = 77.74 \] \[ t^2 = \frac{77.74}{4.9} \approx 15.87 \] \[ t \approx \sqrt{15.87} \approx 3.99 \text{ sec} \] ### **Final Answer:** The closest answer is **3.98 sec**. **So, the correct option is:** - **3.98 sec**