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

Let `t_(0)` be the time -interval of fringe and `O` the origin.
Let the shots collide after time `t` from firing of shot (1). Time for shot (2) `= t - t_(0)`
Shot 1:
`x = 5sqrt(3) cos 60^(@) t = (5sqrt(3))/(2) t` (i)
`y = 10 + (5sqrt(3) sin 60^(@) t - (1)/(2)g t^(2))` (ii)
Shot 2:
`x = 5sqrt(3)(t - t_(0))` (iii)
`y = 10 - (1)/(2)g(t - t_(0))^(2)` (iv)
Equating (i) and (iii), we get
`(5sqrt(3))/(2)t = 5sqrt(3)(t - t_(0)) rArr t = 2t - 2t_(0) rArr t = 2t_(0)`
Equating (ii) and (iv), we have
`10 + (5sqrt(3) sin 60^(@) t - (1)/(2)g t^(2)) = 10 - (1)/(2)g(t - t_(0))^(2)`
`10 + 5sqrt(3)((sqrt(3))/(2))(2 t_(0)) - (1)/(2)g(2t_(0))^(2)`
`= 10 - (1)/(2)g(2 t_(0) - t_(0))^(2)`
`15 t_(0) - 2 t_(0)^(2) = - 5t_(0)^(2) rArr t_(0) = 1 s`
`x = 5sqrt(3)(t - t_(0))`
`= 5sqrt(3) t_(0) = 5sqrt(3) m`
`y = 10 - (1)/(2) g(t - t_(0))^(2)`
`= 10 - (1)/(2)g t_(0)^(2) = 5 m`
Time-interval of firing `= 1 s`
Coordinates of `P = (5sqrt(3), 5 m)`