Q2. A ball is projected with velocity \( 10 \mathrm{~m} / \mathrm{sec} \) at angle of 30 o with the horizontal surface. The time taken by ball to reach the ground is 1 sec 2 sec

When the ball is projected at an angle of 30 degrees, we can use the vertical component of the initial velocity to determine the time it takes to reach the ground. The vertical component can be calculated using \( v_y = v \sin(\theta) \). Here, \( v = 10 \, \text{m/s} \) and \( \theta = 30^\circ \), so \( v_y = 10 \sin(30^\circ) = 5 \, \text{m/s} \). The time of flight until it returns to the ground can be given by the formula \( T = \frac{2 v_y}{g} \), where \( g \approx 9.81 \, \text{m/s}^2 \). Using this, we find \( T = \frac{2 \cdot 5}{9.81} \approx 1.02 \, \text{sec} \), meaning it’s closer to 1 second. In projectile motion, it's key to remember that the time taken to rise to the maximum height is the same as the time taken to fall back down if air resistance is negligible. This symmetry allows for easier calculations, so always look to break down the problem into its vertical and horizontal components for clarity and accuracy!