B

Step 1: Calculate the vertical component of the velocity. The velocity is given as 30 m/s and the angle is 30 degrees.
Vertical velocity, \( v_y = v \sin(30) = 30 \times \frac{1}{2} = 15 \, \text{m/s} \).
Step 2: Use the kinematic equation to find the time to reach the ground. The vertical motion equation is: \( h = v_y t + \frac{1}{2}gt^2 \). In this case, the height \( h = 20 \, ext{m} \) and \( g = 9.8 \, ext{m/s}^2 \). Substituting the values:
\( 20 = (15)t + \frac{1}{2}(9.8)t^2 \).
Step 3: Rearranging gives: \( 0 = 4.9t^2 + 15t - 20 \).
Step 4: Using the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 4.9, b = 15, c = -20 \) gives us the positive root \( t \approx 2.2 \, ext{s} \).
Therefore, Option B is incorrect as it states the ball returns to 20 m after 3 seconds, which is not supported by calculations.