Two guns situated at the top of a hill of height 10m, fire one shot each with the same speed 5(√3)ms^(−1) at some interval of time. One gun fires horizontally and the other fires upwards at an angle of 60° with the horizontal. The shots collide in air at a point P. Find (i) the time interval between the firings and (ii) the coordinates of the hill right below the muzzle and trajectories in x-y plane.


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Question Number 113657 by Ar Brandon last updated on 14/Sep/20

$Two guns situated at the top of a hill of height 10 m, fire one shot each with$ $the same speed 5 \sqrt{3} {ms}^{- 1} at some interval of time . One gun fires horizontally$ $and the other fires upwards at an angle of 60 ° with the horizontal . The shots$ $collide in air at a point P . Find (i) the time interval between the firings and$ $(ii) the coordinates of the hill right below the muzzle and trajectories in x - y$ $plane .$
Commented by Dwaipayan Shikari last updated on 14/Sep/20

$I n t e r v a l o f t i m e 1 s e c o n d$ $c o o r d i n a t e s P (5 \sqrt{3}, 5)$
Commented by Dwaipayan Shikari last updated on 14/Sep/20

$A n I I T - J E E q u e s t i o n$ $I t o o k m a n y t i m e t o s o l v e t h i s . G r e a t q u e s t i o n . I w i l l s o l v e i t l a t e r :)$
Commented by Ar Brandon last updated on 14/Sep/20
Thanks, I found the solution. It's a question from JEE 1996 ��
Commented by Dwaipayan Shikari last updated on 14/Sep/20

$P a r t i c l e A n e e d s t_{1} t i m e t o r e a c h p o i n t P$ $P a r t i c l e B t_{2}$ $s o u p w a r d v e l o c i t y = 5 \sqrt{3} . \frac{\sqrt{3}}{2} = \frac{15}{2} \frac{m}{s} (A t 60 ° a n g l e)$ $A f t e r 2 s e c o n d s i t s w i l l b e a t$ $S = \frac{15}{2} . 2 - 5 {(4)}^{2} = - 5 m (R e l a t i v e h e i g h t)$ $A f t e r 1 s e c o n d o f t h e f i r s t p a r t i c l e t h e s e c o n d p a r t i c l e s t a r t s f l i g h t i n g$ $I t h a s n o v e r t i c a l v e l o c i t y (f r e e f a l l)$ $1 s e c o n d l a t e r I t w i l l c r o s s S = - \frac{1}{2} g {(1)}^{2} = - 5 m$ $I f t h e y c o l l i d e t h e n d i s t a n c e t r a v e l e d i n h o r i z o n t a l$ $= (1) . 5 \sqrt{3} m = 5 \sqrt{3} m$ $S o c o l l i s i o n c o o r d i n a t e P (5 \sqrt{3}, - 5)$ $A n d t i m e d i f f e r e n c e = 1 s e c$
	Answered by Ar Brandon last updated on 14/Sep/20

	$For Bullet A . Let t be the time taken by bullet A to reach P .$ $Verctical motion$ $u_{y} = 0; s_{y} = 10 - y; a_{y} = 10 m / s^{2}; t_{y} = t$ $s_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2}$ $10 - y = {5 t}^{2} . . . (i)$ $Horizontal motion$ $x = 5 \sqrt{3} t . . . (ii)$ $For bullet B$ $Let (t + t^{'}) be the time taken by bullet B to reach P$ $Vertical Motion$ $Considering updward direction negative$ $u_{y} = - 5 \sqrt{3} \sin 60 ° = - 7 . 5 m / s, a_{y} = + 10 m / s^{2}$ $s_{y} = + (10 - y); t_{y} = t + t^{'}, s_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2}$ $\Rightarrow 10 - y = - 7.5 (t + t^{'}) + 5 {(t + t^{'})}^{2} . . . (iii)$ $Horizontal motion$ $x = (5 \sqrt{3} \cos 60 °) (t + t^{'})$ $\Rightarrow 5 \sqrt{3} t + 5 \sqrt{3} t^{'} = 2 x . . . (iv)$ $(ii) in (iv)$ $\Rightarrow 5 \sqrt{3} t + 5 \sqrt{3} t^{'} = 10 \sqrt{3} t \Rightarrow t = t^{'}$ $t = t^{'} in eq . (iii) \Rightarrow y - 10 = 15 t - {20 t}^{2} . . . (v)$ $(i) + (v) \Rightarrow 0 = 15 t - {15 t}^{2} \Rightarrow t = 1 \sec$ $\Rightarrow x = 5 \sqrt{3}, y = 5 . P (5 \sqrt{3}, 5)$
	Commented by Ar Brandon last updated on 14/Sep/20

	Commented by malwan last updated on 14/Sep/20

	$g r e e e e e t$ $t h a n k s a l o t$
	Answered by mr W last updated on 15/Sep/20

	$p o s i t i o n o f f i r s t b u l l e t :$ $x_{1} = u \cos θ (t + Δ t)$ $y_{1} = h + u \sin θ (t + Δ t) - \frac{1}{2} g {(t + Δ t)}^{2}$ $p o s i t i o n o f s e c o n d b u l l e t :$ $x_{2} = u t$ $y_{2} = h - \frac{1}{2} {g t}^{2}$ $u \cos θ (t + Δ t) = u t$ $\Rightarrow t + Δ t = \frac{t}{\cos θ}$ $h + u \sin θ (t + Δ t) - \frac{1}{2} g {(t + Δ t)}^{2} = h - \frac{1}{2} {g t}^{2}$ $\frac{t}{\cos θ} [u \sin θ - \frac{g t}{2 \cos θ}] = - \frac{1}{2} {g t}^{2}$ $\Rightarrow t = \frac{2 u}{g \tan θ} = \frac{2 \times 5 \sqrt{3}}{10 \times \sqrt{3}} = 1 s e c$ $\Rightarrow Δ t = (\frac{1}{\cos θ} - 1) \frac{2 u}{g \tan θ}$ $= (2 - 1) \frac{2 \times 5 \sqrt{3}}{10 \times \sqrt{3}} = 1 s e c$ $x = 5 \sqrt{3}$ $y = 10 - \frac{10}{2} \times 1^{2} = 5$
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