Two guns situated on the top of a hill of height $10$ m fire one shot each with the same speed $5\sqrt{3}$ m/s 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. Taking the origin at the foot of the hill right below the muzzles, the time interval between the firings and the coordinates of P are ($g = 10\ \text{m/s}^2$)
Answer: (A) $1$ s, $(5\sqrt{3}\ \text{m},\ 5\ \text{m})$
Let the horizontal shot be in flight for $t_1$ and the inclined shot for $t_2$ when they meet.
Horizontal positions must match:
$$5\sqrt{3}\,t_1 = 5\sqrt{3}\cos 60°\,t_2 \;\Rightarrow\; t_2 = 2t_1$$
Vertical positions must match ($5\sqrt{3}\sin 60° = 7.5$ m/s):
$$10 - 5t_1^2 = 10 + 7.5t_2 - 5t_2^2 \;\Rightarrow\; -5t_1^2 = 15t_1 - 20t_1^2 \;\Rightarrow\; t_1 = 1\ \text{s}$$
So $t_2 = 2$ s: the inclined gun fires first and the other fires $1$ s later.
P: $x = 5\sqrt{3}(1) = 5\sqrt{3}$ m, $y = 10 - 5(1)^2 = 5$ m.
Solution by Sreeraj P, M.Sc Physics