A projectile is projected at 30° from horizontal with initial velocity 40 ms^-1. The velocity of the projectile at t = 2 s from the start will be: (Given g = 10 m/s²)

Calculation:

​Given: The projectile is projected at an angle of 30° with initial velocity 40 m/s.

We need to find the velocity of the projectile at t = 2 seconds, given g = 10 m/s².

We can resolve the initial velocity into horizontal and vertical components:

Initial horizontal velocity, v_x0 = 40 × cos(30°) = 40 × √3/2 = 20√3 m/s

Initial vertical velocity, v_y0 = 40 × sin(30°) = 40 × 1/2 = 20 m/s

The horizontal velocity remains constant throughout the motion as no horizontal acceleration exists. Thus, at t = 2 seconds, the horizontal component of velocity is:

v_x = 20√3 m/s

The vertical velocity at any time t is given by:

v_y = v_y0 - g × t

v_y = 20 - 10 × 2 = 20 - 20 = 0 m/s

Therefore, at t = 2 seconds, the vertical velocity is zero.

The total velocity of the projectile at t = 2 seconds is the horizontal velocity alone:

v = v_x = 20√3 m/s ≈ 34.64 m/s

So, the velocity of the projectile at t = 2 seconds is approximately 34.64 m/s, which matches Option 1.

Thus, the correct option is Option 1: 20√3 m/s