Question

Two guns situated on the top of a hill of height $10\text{m}$ fire one shot each with the same speed at some interval of time. One gun fires horizontally and other fires upward at an angle of ${60}^{\circ }$ with the horizontal. The shots collide in air at a point P. Then

• A. the time-interval between the shots is 1 sec.
• B. the time-interval between the shots is 2 sec.
• C. the coordinates of the point P is $(5\sqrt{3},5)$ . Take origin of the coordinate system at the foot of the hill right below the muzzle and trajectories in $x–y$ plane.
• D. the coordinates of the point P is $(5,5\sqrt{3})$ . Take origin of the coordinate system at the foot of the hill right below the muzzle and trajectories in $x–y$ plane.

Answer 1

Answer: (A), (C)

The correct options are
A the time-interval between the shots is 1 sec.
C the coordinates of the point P is $(5\sqrt{3},5)$ . Take origin of the coordinate system at the foot of the hill right below the muzzle and trajectories in $x–y$ plane.

Let the shot fired horizontally be denoted by H and the one fired at an angle be denoted by A.
Let the time interval between two shots fired be $\Delta t$ and the time at which collision happens be $t$. Its obvious that in order to have collision between the two shots, the one fired at some angle will be the first one. Let the point at which collision happens be P and its horizontal distance from the hill be $x$.
Time taken by shot H to reach P, $T_H=t-\Delta t$
Time taken by shot A to reach P, $T_H=t$
Using the first equation of motion for horizontal displacement for both the shots.
For shot H, $\begin{array}{}x=v(t-\Delta t)& \text{(1)}\end{array}$
For shot A, $\begin{array}{}x=vcos{60}^{o}t=\frac{v}{2}t& \text{(2)}\end{array}$
Equating (1) and (2),
$v(t-\Delta t)=\frac{v}{2}t$
$\begin{array}{}\Rightarrow t=2\Delta t& \text{(3)}\end{array}$
If the foot of the hill is considered as origin, coordinate of the top of hill will be (0,10). Let the vertical distance of point P from top of the hill be $y$.
Vertical displacement for both the shots($-y$ from top of hill) can be written using 2nd equation of motion.
For shot H,
$\begin{array}{}-y=-\frac{1}{2}g(t-\Delta t{)}^2& \text{(4)}\end{array}$
For shot A,
$\begin{array}{}-y=vsin{60}^o-\frac{1}{2}g{t}^2& \text{(5)}\end{array}$
Equating (4) and (5),
$-\frac{1}{2}g(t-\Delta t{)}^2=vsin{60}^o-\frac{1}{2}g{t}^2$
Since $t=2\Delta t$ and $v=5\sqrt{3}m/s$
$\Rightarrow -\frac{1}{2}g(\frac{t}{2}{)}^2=5\sqrt{3}\frac{\sqrt{3}}{2}-\frac{1}{2}g{t}^2$
Rearranging the above equation, we get
$t=2sec$
Therefore time interval between the shots, $\Delta t=\frac{t}{2}=1sec$
Hence option A is correct.
Rewritting equation (2) for x-coordinate of P,
$x=\frac{v}{2}t=\frac{5\sqrt{3}}{2}\times 2=5\sqrt{3}m$
Rewritting equation (4) for y-coordinate of P,
$-y=-\frac{1}{2}g(t-\Delta t{)}^2$
$y=\frac{1}{2}\times 10\times (2-1{)}^2=5m$
Hence y-coordinate of P, $y=10-5=5m$
Thus coordinates of point P is $(5\sqrt{3},5)$
Hence option C is correct.