AP EAMCET2013PhysicsMotion in Two Dimensions
A particle is projected from the ground with an initial speed of v at an angle of projection . The average velocity of the particle between its time of projection and time it reaches highest point of trajectory is
Options
- Av 2 1+2 ^2
- Bv 2 1+2 ^2
- Cv 2 1+3 ^2
- Dv
Correct answer
C. v 2 1+3 ^2
Step-by-step solution
aligned & We know, average velocity = displacement time & v_ av = ^2+R^2 / 4 T / 2 aligned where, H= maximum height = v^2 ^2 2 g Range R= v^2 2 g Time of flight T= 2 v g Putting the values of Eqs. (ii), (iii) and (iv) in Eq. (i) we have v_ av = v 2 1+3 ^2