Saved Bookmarks
| 1. |
The velocity of a projectile when at its 2 greatest height is sqrt(2/5)of its velocity when at half of its greatest height find the angle of projection |
|
Answer» Solution :Step 1: we know that, velocity of a projectile at HALF of maximum height: `=usqrt((1+cos^(2)theta)/2)` Step 2: given that `u cos theta =SQRT(2/5) xx u sqrt((1+cos^(2)theta)/2)` Squaring on both sides `u^(2) cos^(2)theta =2/5u^(2)(1+cos^(2)theta)/2` `10cos^(2)theta =2 + 2 cos^(2)theta RARR theta=60^(@)` |
|