Question

Two guns situated on the top of a hill of height $10m$ fire one shot each with the same speed $5\sqrt{3}m/s$ at some interval of time. One gas fires horizontally and other fires upwards at an angle of ${60}^o$ with the horizontal. The shots collide in air at point $P$ find
I. The time interval between the firings and
II. The coordinates of the point $P$. Take origin of the coordinate system at the foot of the hill right below the muzzle and trajectories in x-y plane (take $g=10m/s^2$)

Answer 1

Let gun $1$ fire at $t=0sec$ and gun 2 fire at $t=ssec$ After $tsec$

For gun 1

$V_x=5\sqrt{3}m/s{a}_x\Rightarrow x_1=5\sqrt{3}t{V}_y=0,a_y=-g\Rightarrow y_1=0-\frac{1}{2}g{t}^2$

For gun 2

$V_x=5\sqrt{3}\cos 60°,a_x=0\Rightarrow x_2=5\sqrt{3}\cos 60°\left(t_{-}{t}_1\right)V_y=5\sqrt{3}\sin 60°,a_x=-g\Rightarrow y_2=5\sqrt{3}\sin 60°(t-t_1)-\frac{1}{2}g(t-t_1{)}^2$

At collison $x_1=x_2\&y_1=y_2$

$\Rightarrow 5\sqrt{3}t=5\sqrt{3}\frac{1}{2}(t-t_1)$

$\Rightarrow 5\sqrt{3}t=\frac{5\sqrt{3}t}{2}-5\sqrt{3}{t}_1$

$\&-5t^2=\frac{5\sqrt{3}.\sqrt{3}}{2}(t-t_1)-5(t-t_1{)}^2$

Solving we get $t_1=-1sec\&t=1sec$

$\because t_1$ is negative

Gun 2 fires before gun1

Now, $x=5\sqrt{3}t=5\sqrt{3}y=-\frac{1}{2}g{t}^2=-5$

$\therefore$ Collison point $P=(5\sqrt{3},-5)$