A ball is thrown at angle thetha with the horizontal ground with speed u. The maximum height achieved by the particle is H and it is at height h above the ground at times t1 and t2 as shown. The time interval t2 - t1 is equal to

To know the average time traveled from being one point of height to other which is same from the ground determines the parabolic shapes of projectile. Hence given as speed u at angle \theta and time as  and . Horizontal component of the velocity be u cos \theta and the total displacement in travelling from  to  be, 

            uCosQ(t1-t2)

            

=> Total time taken be . (t1-t2)

Therefore the average velocity be, 

Total displacement /total time 

Va = uCosQ(t1-t2) / (t1-t2)

Va = u CosQ