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.

Answer

(a) `1sec`, (b) `(5sqrt(3)m, 5m)`

Explanation

For the horizontally
`zx = 5sqrt(3)t_(1), …(1)`
For fire at `60^(@)`
`x = 5sqrt(3) cos 60^(@) t_(2) ..(2)`
`x = (5sqrt(3))/(2) t_(2)`
From (1) and (2)
`t_(2) = 2t_(1) …(3)`
For fire horizontally
`y = 10- (1)/(2) gt_(1)^(2) ...(4)`
For fore at an angle `60^(@)`
`y = 10 +5 sqrt(3) sin 60^(@) t_(2) - (1)/(2) gt_(2)2 ...(5)`
From (4) and (5)
`10 +5sqrt(3) xx (sqrt(3))/(2)t_(2) - (1)/(2) g t_(2)^(2) = 10 - (1)/(2) g t_(1)^(2)`
`(15)/(2) t_(2) - 5t_(2)^(2) +5t_(1)6(2) = 0 ...(6)`
from (3) & (6) `15t_(1) - 20t_(1)^(2) +5t_(1)^(2) = 0`
`t_(1) = 1sec` & `t_(2) = 2sec`
(a) Time interval between firing `t_(2) - t_(1) = 1sec`.
(b) `x = 5sqrt(3)t_(1)`
`= 5sqrt(3)`
`y = 10- (1)/(2) gt_(1)^(2) rArr y = 5m`