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

Method 1 : Foe bullet A. Let `t` be the time taken by bullet `A` to reach `P`
Vertical motion : `u_y = 0, S = ut + (1)/(2) at^2,S_y = -(10 -y)`
`a_y = -10 m s^-2 rArr 10 - y = 5 t^2` …(i)
Horizontal motion : `x = 5 sqrt(3) t` …(ii)
For bullet B. Let `(t + l)` be the time taken by bullet `B` to reach `P`.
Vertical motion : `uarr + u_y = + 5 sqrt(3) sin 60^@ = + 7.5 m s^-1`
`S = ut + (1)/(2) at^2 , a_y = -10 m s^-2`
`y - 10 = 7.5 (t + t') - 5 (t + t')^2` ...(iii)
Horizontal motion : `x = (5 sqrt(3) cos 60^@)(t + t')`
`rArr 5 sqrt(3) t + 5 sqrt(3) t' = 2x` ...(iv)
Substituting the value of `x` from (ii) in (iv), we get
`5 sqrt(3) t + 5 sqrt(3) t' = 10 sqrt(3)) t rArr t = t'`
(a) Putting `t = t'` in (iii),
`y - 10 = 15 t - 20 t^2` ...(v)
Adding (i) and (v), `0 = 15 t - 15 t^2 rArr t = 1 s`
So time interval between the firings is `t' = 1 s`
(b) Putting `t - 1` in (ii), we get `x = s sqrt(3) m`.
Putting `t = 1` in (i), we get `y = 5`.
Therefore, the coordinates of point
`P are (5 (sqrt(3) , 5))` in meters.
Method 2 : We take the point of firing as origin and `x - and y` - axis as shown in (Fig. 5.174). Equation of trajectory of the projectile is
`y = x tan theta - (g x^2)/(2 v_i^2 cos^2 theta)`
For gun 1,`theta = 60^@`. Therefore,
`y = x tan 60^@ - (g x^2)/(2 v_i^2 cos^2 60^@)= x sqrt(3) - (2 g x^2)/(v_i^2)`
For gun 2, `theta = 0^@`. Therefore,
`y = (- g x^2)/(2 v_i^2)`
Two shots collide at point `P`. Therefore, their coordinates must be same.
`(- g x^2)/(2 v_i^2) = x sqrt(3) - (2 g x^2)/(2 v_i^2)`
`x sqrt(3) = (2 g x^2)/(v_i^2) - (g x^2)/(2 v_i^2) = (3 g x^2)/(2 v_i^2)`
`x = 0` and `x = (2 v_i^2)/(sqrt(3 g) = (2(5 sqrt(3))^2)/(sqrt(3 (10))) =5 sqrt(3)`
`y = (-g x^2)/(2 v_i^2) = (-10(5 sqrt(3))^2)/(2(5 sqrt(3))^2) = -5 m`
If the origin is assigned at ground, the coordinates of point `P` will be `(5 sqrt(3)) m, 5 m)`.
Now we consider the x - component of displacement for both the shots.
Gun 1 : `x = 5 sqrt(3) m = v_i t (5 sqrt(3) ms^-1) t`
or `t_1 = 1 s`
Gun 2 : `x = 5 sqrt(3)m = v_i cos 60^@ t_2 = (5 sqrt(3))/(2) t_2`
or `t_2 = 2 s`
Time interval between two shots, `Delta t = t_2 -t_1 = 1 s`.
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