Two guns situated at the top of a hill of height `10 m` fire one shot each with the same speed `5sqrt(3) m//s` at some interval of time. One gun fires horizontal and the other fores upwards at an angle of `60^(@)` with the horizontal. Two shots collide in air at a poit `P`. Find (i) time-interval between the firing and (ii) coordinates of the point `P`. Take the origin of coordinates system at the foot of the hill right below the muzzle and trajectorise in the `x-y` plane.

Options

A. `(5 m,5m)`
B. `(5sqrt3 m,5sqrt3m)`
C. `(5sqrt3 m,5m)`
D. `(5 m,5sqrt3m)`

Answer

C

Explanation

Let gun `1` and gun `2` to be fired at an interval `Deltat`,such that `t_(1)=t_(2)+Deltat---(1)`
Where `t_(1)` and `t_(2)` are the respective times taken by the two shots to reach point `P`.
For gun 1:
x-component y-component
`{:("x-component",,"y-component"),(x-x_(i)=v_(i) cos 60^(0)t_(1),,y-y_(i)=v_(i) sin 60^(0)t_(1)-(1)/(2)g t_(1)^(2)),(or x=x_(i)+(1)/(2)v_(i)t_(1),,y=y_(i)+(sqrt(3))/(2)v_(i)t_(1)-(1)/(2)g t_(1)^(2)):}`
For gun 2 :
`{:("x-component",,"y-component"),(x-x_(i)=v_(i) cos 60^(0)t_(2),,y-y_(i)=v_(i) sin 0^(0)t_(2)-(1)/(2)g t_(1)^(2)),(or x=x_(i)+v_(i)t_(2),,y=y_(i)-(1)/(2)g t_(2)^(2)):}`
(a) Now we can equate `x-` and y-coordinates of shots
i.e.,`1/2v_(i)t_(i)=v_(i)t_(2) or t_(1)=2t_(2)` and `sqrt3/2v_(i)t_(1)-1/2"gt"_(1)^(2)=-1/2"gt"_(2)^(2)`
or `sqrt3/2v_(i)t_(1)+1/2g(t_(2)^(2)-t_(1)^(2))=0` On substituting `t_(1)`
from eqn. (2) into eqn. (3), we get
`sqrt3/2v_(i)(2t_(2))+1/2g(-3t_(2)^(2))=0 or t_(2)(sqrt3v_(i)-3/2"gt"_(2))=0`
or `t_(2)=0` and `t_(2)=2/sqrt3v_(i)/g=2/sqrt3xx((5sqrt3)/10)=1s`
Therefore, `t_(1)=2t_(2)=2(l)=2s`
`Deltat=t_(1)-t_(2)=2-1=1s`
(b) The coordinates of `P` at which the two shots collide are `x=x_(i)+v_(i)t_(2)=0+(5sqrt3)(1)=5sqrt3m`
and `y=y_(i)-1/2"gt"_(2)^(2)=10-1/2(10)(1)^(2)=5m`