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

A particle is projected from the ground with an initial speed of v at

1.	A particle is projected from the ground with an initial speed of v at a angle theta with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is
Answer» Solution :Average VELOCITY `=("displacement")/("time")` `V_(av)=(sqrt(H^(2)+(R^(2))/4))/(T/2)`…..i Here H=maximum height `=(V^(2)sin^(2)THETA)/(2g)` R=range `=(v^(2)sin 2 theta)/G` and `T=` time of FLIGHT `=(2vsin theta)/g` Substituting i (i) we get `v_(av)=v/2sqrt(1+3 cos^(2)theta)`