InterviewSolution
Saved Bookmarks
| 1. |
A projectile is thrown with speed u making angle `theta` with horizontal at `t=0`. It just crosses the two points at equal height at time `t=1` s and `t=3` sec respectively. Calculate maximum height attained by it. `(g=10 m//s^(2))` |
|
Answer» Correct Answer - 20 m `u sin thetaxx1-1/2 g(1)^(2)=u sin thetaxx3-1/2xxgxx(3)^(2)` `2u sin theta=40 rArr u sin theta=20 m//s` Max. height `=(u^(2)sin^(2) theta)/(2g)=(20xx20)/2-=20 m` |
|