Saved Bookmarks
| 1. |
A bomb initially at rest at a height of 40 m above the ground suddenly explodes into two identical fragments. One of them starts moving vertically downwards with an initial speed of 10 m/s. If acceleration due to gravity is 10 m//s^2, the separation between the fragments, 2 seconds after the explosion is |
|
Answer» Solution :Initial relative velocity `u_(rel) =10-(-10) = 20 m//s` Relative acceleration `a_("rel") =g -g=0` After t=2 sec, relative separation After t=2 sec, relative separation `S_("rel") =u_(rel) t + 1/2 a_(rel)t^(2) =(20 xx 2) + 0= 40 m` |
|