Saved Bookmarks
| 1. |
A particle is projected from level ground its kineitc energy K changes due to gravity so that `K_("max")//K_("min")=9`. The ratio of the range to the maxir, un height attained during its fight isA. `sqrt(2)`B. `4sqrt(2)`C. 1.5D. none |
|
Answer» `KE_("max")=(1)/(2)mv^(2)` (at the point of projection). `KE_("min")=(1)/(2) mv^(2)cos ^(2)theta` `(v^(2))/(v^(2)cos^(2)theta)=9 rArr cos theta=(1)/(3)` as `R=(v^(2)sin2 theta)/(g), H=(v^(2)sin^(2) theta)/(2g)` `R//H=4 cot theta` |
|