A projectile is thrown in the upward direction making an angle of (60^ ) with the horizontal with a velocity of (140 ~ms ⁻¹ ). Then the time after which its velocity makes an angle (45^ ) with the horizontal is (Acceleration due to gravity, (g=10 ~ms ⁻² ) )

Step-by-step solution

Given, angle of projection, ( =60^ ) and velocity, (u=140 ~ms ⁻¹ ) This velocity have divided into two components, horizontal component, (u_x=u 60^ =140 60^ ) and vertical component, (u_y=u 60^ =140 60^ ) Let after time (t ), the inclination of particle with horizontal be (45^ ) and at time (t ) velocity along (x=v_x ) and along (y=v_y ). Now, ( aligned 45^ & = v_y v_x & =v_x=v_y aligned ) Since, the horizontal component of velocity remains constant, i.e., ( aligned v_x & =u_x=140 60^ and v_y & =u_y-g t 140 60^ & =