Saved Bookmarks
| 1. |
Find the angle of projection for which the horizontal range and the maximum Height are equal. |
|
Answer» SOLUTION :`(mu^2sin2theta)/G=(mu^2sin^2theta)/(2G)sin2theta=sin^2theta/22sinthetacostheta=(sin^2theta)/2tantheta=4 0=75.58.` |
|