Question

Two guns situated on the top of a hill of height $15\text{m}$ fire one shot each with the same speed $5\text{m/s}$ with some time interval between the firings. One gun fires horizontally and other fires upwards at an angle of ${45}^{\circ }$ with the horizontal. The shots collide in air at point P. Then

• A. $P(x,y)=p(5,10)$
• B. $P(x,y)=p(10,5)$
• C. Time interval between firings is $0.8\text{s}$
• D. Time interval between firings is $0.4\text{s}$

Answer 1

Answer: (A), (D)

The correct options are
A $P(x,y)=p(5,10)$
D Time interval between firings is $0.4\text{s}$

Given, speed of shots $=5\text{m/s}$
Height of hill $=15\text{m}$
Angle of fire of one shot $={45}^{\circ }$ with horizontal
Angle of fire of other shot $=0^{\circ }$ with horizontal
We know that, equation of trajectory of projectile is given by, $Y=\text{X tan}\theta -\frac{g{X}^2}{2u^{2}co{s}^2\theta }$
[origin of co- ordinate system for this equation is at point of projection]
For shot 1,
$Y=X\times 1-\frac{10X^2}{2\times 5^2\times \frac{1}{2}}$
$Y=X-\frac{2X^2}{5}----\left(1\right)$
Similarly, of shot 2
$Y=X\times 0-\frac{10X^2}{2\times 5^2\times 1}$
$Y=\frac{-X^2}{5}----\left(2\right)$
As both shots collide in air at a point P then its x and y co-ordinates should be same.
From (1) and (2)
$X-\frac{2X^2}{5}=\frac{-X^2}{5}$
$\Rightarrow X-\frac{X^2}{5}=0$
$\Rightarrow X(1-\frac{X}{5})=0$
$\Rightarrow X=0$ and $X=5$ ----(3)
on putting (3) in (2)
$Y=-5$

Therefore wrt original coordinate system with origin at bottom of the tower,
co - ordinates of P is $x=5\text{m}$ and $y=10\text{m}$

Time taken by shot fired at ${45}^{\circ }$ with horizontal to reach point $P,\text{is}{t}_1=\frac{5}{5\text{cos}{45}^{\circ }}=\sqrt{2}=1.4\text{s}$ [Horizontal motion]
Time taken by shot fired at horizontal to reach point $P,\text{is}{t}_2=\frac{5}{5}=1\text{s}$ [Horizontal Motion]
$\therefore$ Time internal between firings for collision to happen is $=t_1-t_2=1.4-1=0.4s$