Saved Bookmarks
| 1. |
A particle is projected from the ground with an initial speed of u at an angle of projection theta. The average velocity of the particle reaches highest point of trajectory is |
Answer» Solution : `v_("AVG") =(VECV + vecu)/2 = (u cos thetahati +(u cos thetahati + u SINTHETA))/2` `v_(AV) =v/2sqrt(1+3 cos^(2)theta)` |
|