A particle is projected from the ground with an initial speed of 20 3 m/s at an angle of 60^ above the horizontal. The time interval between the two instants (one during ascent and one during descent) when its velocity vector makes an angle of 30^ with the horizontal is (Take g = 10 m/s ^2 )

Step-by-step solution

The initial components of velocity are: u_x = u 60^ = 20 3 1 2 = 10 3 m/s u_y = u 60^ = 20 3 3 2 = 30 m/s The horizontal component of velocity remains constant, so v_x = 10 3 m/s at all times. When the velocity vector makes an angle of 30^ with the horizontal, the vertical component v_y satisfies: 30^ = |v_y| v_x 1 3 = |v_y| 10 3 |v_y| = 10 m/s During ascent, v_y = +10 m/s . Using v_y = u_y - gt₁ : 10 = 30 - 10t₁ t₁ = 2 s During descent, v_y = -10 m/s . Using v_y = u_y - gt₂ : -10 = 30 - 10t₂ t₂ = 4 s The time inte