Question

Two guns, situated on top of a hill of height ,10 m, fire one shot ench with the same speed $5\sqrt{3}m{s}^{-1}$ at some interval of time. One gun fires horizontally and the other fires upward at an angle of ${60}^o$ with the horizontal. The shots collide in air at point P. Find (a) the time interval between the fringes and (b) the coordinates of point P. Take the origin of the coordinate system at the foot of the hill right below the muzzle and trajectories in the x-y plane.

Answer 1

Method 1:

For bullet A. Let t be the time taken by bullet A to reach P.

Vertical motion: $u_y=0,S=ut+\frac{1}{2}a{t}^2,S_y=-(10-y)$

$a_y=-10m{s}^{-2},\Rightarrow 10-y=5t^2$ ...(i)

Horizontal motion: $x=5\sqrt{3}t$ ...(ii)

For bullet B. Let $(t+l)$ be the time taken by bullet B to reach P.

Vertical motion: $\uparrow +u_y=+5\sqrt{3}sin{60}^o=+7.5m{s}^{-1}$

$S=ut+\frac{1}{2}a{t}^2;a_y=-10m{s}^{-2}$

$y-10=7.5(t+t^{\prime })-5(t+t^{\prime }{)}^2$ ....(iii)

Horizontal motion: $x=\left(5\sqrt{3}cos{60}^o\right)(t+t^{\prime })$

$\Rightarrow 5\sqrt{3}t+5\sqrt{3}{t}^{\prime }=2x$ .....(iv)

Substituting the value of x from (ii) in (iv), we get

$5\sqrt{3}t+5\sqrt{3}{t}^{\prime }=10\sqrt{3}t$

$\Rightarrow t=t^{\prime }$

a. Putting $t=t^{\prime }$ in (iii),

$y-10=16t-20t^2$ ..(v)

Adding (i) and (v),

$0=15t-15t^2\Rightarrow t=1s$

So time interval between the firings is $t^{\prime }=1s$

b. Putting $t=1$ in (ii), we get $x=5\sqrt{3}m.$

Putting $t=1$ in (i), we get $y=5$. Therefore, the coordinates of point $P$ are $(5\sqrt{3},5)$ in meters.