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.

Explanation

As shown in Fig. 5.43, from equation of motion
`s=-s_(0)+ur+(1)/(2)at^(2)` for first gun,

`x_(1)=5sqrt(3)t_(1)` ......(i)
and `y_(1)=10-(1)/(2)gt_(1)^(2)` .........(ii)
While for the second gun
`x_(2)=5sqrt(3)cos 60^(@)t_(2)` ........(iii)
and `y_(2)=10+5sqrt(3) sin 60^(@)t_(2)-(1)/(2)gt_(2)^(2)`
For collision
`x_(1)=x_(2)` and `y_(1)=y_(2)`
`i.e., 5sqrt(3)t_(1)=5sqrt(3)(1)/(2)t_(2)` and `10-(1)/(2)gt_(1)^(2)=10+(15)/(2)t_(2)-(1)/(2)gt_(2)^(2)`
i.e., `t_(2)=2t_(1)` and `g(t_(2)^(2)-t_(1)^(2))=15t_(2)`
Which on simplification gives `t_(1)=1` sec and `t_(2)=2` sec.
So, the interval between the firings =2-1=1 sec and substituting `t_(1)=1` s and g=10 `m//^(2)` in Eqns. (i) and (ii) co-ordinates of point P will be `(5sqrt(3),5)m`.