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, C

Explanation

For Bullet `A`. Let `t` be the time taken by bullet `A` to reach `P` . Vertical motion
` u_(y) = 0 , s_(y) = 10 - y , a_(y) = 10 m//s^(2) , t_(y) = t `
` s_(y) = u _(y)t + 1/2 a_(y)t^(2) `
`10 - y = 5 t^(2) ` … (i)
Horizontal motion ` x = 5 sqrt(3)t ` ... (ii)
For bullet `B`.
Let `( t + t' )` be the time taken by bullet `B` to reach `P`
Vertical Motion
Let us consider upward direction negative and downward positive . Then ,
` u_(y) = -5sqrt(3) sin 60(@) = -7.5 m//s , a_(y) = + 10 m//s ^(2)`
` s_(y) = + (10 - y) , t_(y) = t + t' , s_(y) = u_(y)t + 1/2 a_(y)t^(2)`
` 10-y = - 7.5(t + t') + 5 (t + t')^(2)` .... (iii)
Horizontal motion
` x = (5sqrt(3) cos 60(@)) ( t + t')`
`rArr 5 sqrt(3) t + 5 sqrt(3) t' = 2 x` ....(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'`
Putting ` t = t' ` in eq. (iii)
` y-10 = 15t - 20t ^(2) ` ....(v)
Adding (i) and (v)
` 0 = 15 t - 15 t^(2) rArr t = 1 sec`.
(ii) Putting ` t = 1` in eq. (ii) we get ` x = 5 sqrt(3) `
putting `t = 1 ` in eq. (i) , we get ` y= 5 `
Therefore , the coordinates of point `P` are `( 5sqrt(3) , 5) in metres` .