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, $(5\sqrt{3}\ \text{m},\ 5\ \text{m})$
Concept. Two shots collide when they are at the same point at the same instant, so both their $x$ and $y$ coordinates must match. Because they are fired at different times, give each its own flight time ($t_1$ for the horizontal shot, $t_2$ for the inclined shot) and solve the two equations.
Step 1. Components of the inclined shot. With $u=5\sqrt3$ m s$^{-1}$ at $60^\circ$:
$$u\cos60^\circ=\frac{5\sqrt3}{2}\ \text{m s}^{-1},\qquad u\sin60^\circ=5\sqrt3\cdot\frac{\sqrt3}{2}=7.5\ \text{m s}^{-1}$$
Step 2. Match the $x$-coordinates. Both start at $x=0$:
$$5\sqrt3\,t_1=\frac{5\sqrt3}{2}\,t_2\ \Rightarrow\ t_2=2t_1$$
Step 3. Match the $y$-coordinates. Both start at $y=10$ m:
$$10-5t_1^2=10+7.5t_2-5t_2^2$$
Substitute $t_2=2t_1$:
$$-5t_1^2=15t_1-20t_1^2\ \Rightarrow\ 15t_1^2=15t_1\ \Rightarrow\ t_1=1\ \text{s},\ t_2=2\ \text{s}$$
Step 4. Time interval and point P. The inclined shot is in the air $1$ s longer, so it is fired first and the interval is $1$ s.
$$x_P=5\sqrt3(1)=5\sqrt3\ \text{m},\qquad y_P=10-5(1)^2=5\ \text{m}$$
Answer. $1$ s, $(5\sqrt3\ \text{m},\ 5\ \text{m})$ (Option A). Check with the inclined shot: $y=10+7.5(2)-5(4)=5$ m.
Solution by Sreeraj P, M.Sc Physics
