Two seconds after projection, a projectile is travelling in a direction inclined at 30^ to the horizontal. After one more sec, it is travelling horizontally, the magnitude and direction of its velocity are

Step-by-step solution

Let in 2 s body reaches upto point A and after one more sec upto point B . Total time of ascent for a body is given 3 s array lrl i.e., & & t & = u g =3 & u & =10 3 & u & =30...(i) array Horizontal component of velocity remains always constant u =v 30^ ...(ii) For vertical upward motion between point O and Av 30^ =u -g 2 [ Using v=u-g t] aligned & v 30^ =30-20 [A s u =30] & v=20 ~m / s aligned Substituting this value v , we get in Eq. (ii) u =20 30^ =10 3 ...(iii) From Eqs. (i) and (iii) u=20 3 and =60^