Saved Bookmarks
| 1. |
A particle is projected with speed `10 m//s` at angle `60^(@)` with the horizontal. Then the time after which its speed becomes half of initial.A. `(1)/(2)sec`B. `1sec`C. `sqrt(3//2)sec`D. `sqrt(3//2)sec` |
|
Answer» Correct Answer - D `u cos60^(@)=5,V_(y)=usin 60^(@)-10t` `V_(2)=(u sin 60^(@)-10t)_(2)+(u cos 60^(@))` `(u^(2))/(4)=(u(sqrt(3))/(2)-10t)^(2)+(u^(2))/(4)` `rArr 10t=(10 sqrt(3))/(2) rArr t=(sqrt(3))/(2)`. |
|