A particle is projected from the ground with an initial speed of v at an angle θ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is

detailed solution

Correct option is C

The average velocity is given by vav= displacement  time The situation of the particle at the highest point is shown in fig. (4).We know thatMaximum height, H=v2sin2⁡θ2g        ….(i)Range,      R=v2sin⁡2θg           ….(ii)and Time of flight, T=2vsin⁡θg    ….. (iii)Now using triangle OPQ, we havevav=OP(T/2)=2(OP)Tor         vav=2H2+R2/41/2T    …(iv)substituting the values in eq. (iv) from eqs. (i), [ii) and (iii), we getvav=v21+3cos2⁡θ