A Bomber flying upward at an angle of `53^(@)` with the vertical releases a bomb at an altitude of 800 m. The bomb strikes the ground 20 s after its release. Find : [Given `sin 53^(@)=0.8, g=10 m//s^(2)`]
(i) The velocity of the bomber at the time of release of the bomb.
(ii) The maximum height attained by the bomb.
(iii) The horizontal distance tranvelled by the bomb before it strikes the ground
(iv) The velocity (magnitude & direction) of the bomb just when it strikes the ground.

Answer

(i) `100 m//s` (ii) 980 m (iii) 1600 m (iv) `(80hat(i)-140 hat(j))`

Explanation

`20=(0.6 v)/g+sqrt(2/gxx[((0.6v)^(2))/(2g)+800])` ...(i)
(i) By solving equation (i), we get `v=100 m//s`
(ii) Maximum height:
`=800+((0.6v)^(2))/(2g)=800+((0.6xx100)^(2))/20=980 m`
(iii) horizontal distance
=Horizontal velocity `xx`time of flight
`=100 cos 37^(@)xx20=1600m`
(iv) horizontal component
`v_(H)=u_(H)=100 cos 37^(@)=80 m//s`
`v_(v)=u_(v)-10xx20=100 sec 37^(@)-200`
`=140 m//s`
`:. v_("strike")=80hat(i)-140hat(j), |vec(v)|=sqrt(80^(2)+140^(2))`