Question

Two guns situated at the top of a hill of height $10m$, fire one shot each with the same speed of $5\sqrt{3}m{s}^{-1}$ at some interval of time. One gun fires horizontally and other fires upwards at an angle of ${60}^0$ with the horizontal. The shots collide in mid air at the point $P$. Taking the origin of the coordinate system at the foot of the hill right below the muzzle, trajectories in x-y plane and $g=10{ms}^{-2}$ then

• A. they collide at P whose coordinates are given by $(5,5\sqrt{3})m$
• B. time difference between firing of shots is $2s$
• C. time difference between firing of shots is $1s$
• D. they collide at P whose coordinates are given by $(5\sqrt{3},5)m$

Answer 1

Answer: (C), (D)

The correct options are
C time difference between firing of shots is $1s$
D they collide at P whose coordinates are given by $(5\sqrt{3},5)m$

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

For B $\to u_h=5\sqrt{3}\cos {60}^{\circ }=\frac{5\sqrt{3}}{2}m/s;$
$u_v=5\sqrt{3}\sin {60}^{\circ }=7.5m/s$

Ball B has been fired before A by $T$ sec.
Time taken by A $=t$

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

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

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

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

Co-ordinates $\to R=5\sqrt{3}$
$H=\frac{1}{2}\times 10\times {\left(1\right)}^2=5$

thus (C) and (D) are true.