A shell is fired from a point O at an angle of `60^(@)` with a speed of `40 m//s` & strikes a horizontal plane throught O, at a point A. The gun is fired a second time with the same angle of elevation but a different speed v. If it hits the target which starts to rise vertically from A with a constant speed `9sqrt(3) m//s` at the same instant as the shell is fired, find v.

Answer

`50 ms^(-1)`

Explanation

Range `(OA)=(u^(2) sin 2 theta)/(g)=(1600 xx sqrt(3))/(10 xx 2)=80 sqrt(3)`
`h = 80 sqrt(3) xx tan 60^(@) = (10 xx 80 xx 80 xx 3)/(2 xx v^(2) cos 60^(@))`
Time to strike `rArr v cos 60^(@) xx t = 80 sqrt(3)`
`rArr t =(80 sqrt(3) xx2)/(v xx1) =(10 sqrt(3))/(v)`
`h=9 sqrt(3) xx (160 sqrt(3))/(v) (480 xx 9)/(v) = (240^(2) - 38400)/(v^(2))`
`v^(2) -1600 - 18v =0`
`v =(18 pm sqrt(324)+6400)/(2)`
`rArr v =50 m//s`.