Question

Two guns situated on the top of a hill of height $10m$, each fired shots with the same speed $5\sqrt{3}m/s$ at some interval of time. One gun fires horizontally and other fires upwards at an angle of ${60}^{\circ }$ with the horizontal, the shots collide in air at a point $P$. Find:
(i) the time interval between the firings and
(ii) the cooridnates of the point $P$. Take origin of the coordinate system at the foot of the hill right below the muzzle and trajectories in $x-y$ planes.

Answer 1

REF.Image.

For $A\to u_h=5\sqrt{3}m/s;u_v=0$

For $B\to u_h=5\sqrt{3}cos{60}^{\circ }=5\sqrt{3}m/s$

$u_v=5\sqrt{3}sin{60}^{\circ }=7.5m/s^2$

Ball B has been field before A by T sec.

Time taken by A = t.

Equating horizontal distance:

$R_A=R_B\Rightarrow 5\sqrt{3}\times t=\frac{5\sqrt{3}}{2}\times (T+t)\Rightarrow T=t$

Equating vertical distance

$H_A=H_B\Rightarrow \left(0\right)t-\frac{1}{2}\times 10\times t^2=\left(7.5\right)(T+t)-\frac{1}{2}\times 10\times (t+T{)}^2$

$\Rightarrow -5t^2=\frac{15}{2}\times 2t-5\times 4t^2$

$\Rightarrow 15t^2=15t\Rightarrow t=1$

co - ordinates $\to R=5\sqrt{3}$

$H=\frac{1}{2}\times 10\times (1{)}^2=5$