Question Statement:-

An enemy jet is flying at a constant height of $250\text{ m}$ with a velocity of $500\text{ m/s}$. The fighter jet passes over an anti-aircraft gun that can fire at any time and in any direction with a speed of $100\text{ m/s}$. Determine the time interval during which the fighter jet is in danger of being hit by the gun bullets.

Source:- Mechanincs for JEE (Main & Advanced) (Vol.-1) - Er. Anurag Mishra

My solution:-

As we know the equation of trajectory of a projectile projected at an angle $\theta$ under the influence of gravitational force is given by

$$y=x\tan{\theta}-\dfrac{gx^2}{2u^2}(1+\tan^{2}{\theta}) \tag{1} $$

where $x$ and $y$ represent the cartesian coordinates.

Now, consider $t=0\text{ sec}$ as the point in time when the anti-aircraft gun fires a bullet to hit the the jet plane. So, to hit the aircraft the bullet must have the same coordinates as that of the jet plane at the time when the bullet reaches an altitude same as that of the jet plane, i.e. $250\text{ m}$. So, lets assume after $t$ seconds the bullet reaches a height of $250\text{ m}$. At that time the coordinates of the jet plane would be $(500t,250)$, so the $x-$coordinate of the bullet will be $x=500t$.

So, on substituting $u=100\text{ m/s}; g=10\text{ m/$s^2$}; y=250\text{ m}; x=500t$ in equation $(1)$, we get

$$\begin{aligned} &250=(500t)\tan{\theta}-\dfrac{g(500t)^2}{2\times{100}^2}(1+\tan^{2}{\theta}) \\ \implies &\dfrac{g(500t)^2}{2u^2}\tan^{2}{\theta}-(500t)\tan{\theta}+\left(250+\dfrac{g{(500t)}^{2}}{2{u}^{2}}\right)=0 \end{aligned}$$

Now, for the such a trajectory to exist $D\ge0$

$$\begin{aligned} \therefore \qquad(500t)^2-4\dfrac{g(500t)^2}{2u^2}\left(250+\dfrac{g(500t)^2}{2u^2}\right)\ge0\\ \implies t^2(2-t^2)\ge0 \end{aligned}$$

The interval for $t$ is $t \in \left[-\sqrt{2},\sqrt{2}\right]$ ,but as $t\gt 0$, so $t\in (0,\sqrt{2}]$.

So, the jet plane is in danger of being hit by the anti-aircraft gun for an interval of $\sqrt{2}$ seconds

Solution of book:-

The equation of trajectory of bullets is $$y=x\tan{\theta}-\dfrac{gx^2}{2u^2}(1+\tan^{2}{\theta}) \tag{1}$$

danger zone according to book For a given value of $x$, maximum $y$ can be determined from

$$\dfrac{dy}{d(\tan{\theta})}=x-\dfrac{gx^2}{u^2}(\tan{\theta})=0 \\ \implies \tan{\theta}=\dfrac{u^2}{gx}$$

On substituting the expression for $\tan{\theta}$ in equation $(1)$, we get

$$\begin{aligned} &y_{max}=\dfrac{u^2}{g}-\dfrac{gx^2}{2u^2}\left[1+\dfrac{u^4}{g^2x^2}\right] \\ \implies & y_{max}= \dfrac{u^2}{2g}-\dfrac{gx^2}{2u^2} \end{aligned}$$

The shell can hit an area defined by

$$y\le \dfrac{u^2}{2g}-\dfrac{gx^2}{2u^2}$$

On substituting numerical values, $y=250 m; u=100 m/s; g=10 m/s^2$, we get

$$\dfrac{x^2}{2000}\le 250 \implies -500\sqrt{2}\le x \le 500\sqrt{2}$$

The fighter jet, can travel $1000\sqrt{2}$ m while it can be hit. So, the plane is in danger for a period of $$\dfrac{1000\sqrt{2}}{500}=2\sqrt{2}\text{sec}$$

My deal with the question:-

  1. Why does my answer and the books answer differ, what am I missing.
  2. Why does the book's solution find out the maximum $y$ for a particular $x$(which I think is the position of the jet at some time as explained in my solution, correct me if I am wrong).
  3. Why is it that the projectile can hit only the area defined in the solution, did it mean to say that only in that area the shell can hit the jet plane, if that is so then still I have the doubt why only that region.
  4. Lastly, what could have been the inspiration of the author to come up with this solution

More elegant solution are always welcome.

Edit 1:- I know this question has been asked here, but as the OP did not provide any work so it has been put on hold, hence I provided my work and also, this is not a homework, I am solving it on my own.

Edit 2:-

As pointed out by all of the solvers, I had done a mistake in determining the reference time for the jet, so to correct that I tried the following.

Let the position of the jet in the cartesian plane be $(x+500t,250)$, where $-\infty\le x \le\infty$ and $\Delta t$ represents the time elapsed after firing the bullet from the gun and the bullet hitting the jet(as I could not think of anything good which could also tell that $t$ can be negative if we take $t=0$ as the reference time, so instead I defined $\Delta t$). On substituting these values into the equation of trajectory of the projectile we get,

Equation of trajectory of a projectile projected at an angle $\theta$ under the influence of gravitational force is given by

$$y=x\tan{\theta}-\dfrac{gx^2}{2u^2}(1+\tan^{2}{\theta}) \tag{1}$$

On substituting $(x+500t,250)$, in equation $(1)$, we get

$$\begin{aligned} & 250=(x+500t)\tan{\theta}-\dfrac{g(x+500t)^2}{2u^2}(1+\tan^{2}{\theta}) \\ \implies & \dfrac{(500t+x)^2g}{2u^2}\tan^2{\theta}-(500t+x)\tan{\theta}+\left(250+\dfrac{(500t+x)^2g}{2u^2}\right)=0 \\ \implies & \dfrac{(500t+x)^2}{2000}\tan^2{\theta}-(500t+x)\tan{\theta}+\left(\dfrac{5\times10^5+(500t+x)^2}{2000}\right)=0 \end{aligned}$$

Now, for such a trajectory to exist $\tan{\theta}\in \mathbb{R}$, so $D\ge0$

$$\begin{aligned} \therefore \qquad &(x+500t)^2-4\left(\dfrac{(500t+x)^2}{2000}\right)\left(\dfrac{5\times10^5+(500t+x)^2}{2000}\right) \ge 0 \\ \implies & (500t+x)^2(5\times10^5-(500t+x)^2)\ge 0 \end{aligned}$$

Now , let $(500t+x)=p$, then the above inequality becomes

$$\begin{aligned} &p^2(5\times10^5-p^2)\ge0 \\ \implies &-500\sqrt{2}\le p\le 500\sqrt{2} \end{aligned}$$

As we can see that $p=(500t+x)$ represents the $x$-coordinate of the jet, so we get that the jet is in danger as represented in the above inequality for $\boxed{2\sqrt2}$ seconds.

Please do tell me if I had done any mistakes, and seems like the solution of the book is way more intuitive and short.


UPDATED Sammy Gerbil rightly pointed out that my answer was wrong - and I apologize. So here is the "real" answer...

Let's start with the following diagram:

enter image description here

It is a diagram of the possible trajectories of the cannonball, fired at different angles. In green is the "critical" angle - the one that would just reach a height of 250 m. But as you can see - that is NOT the angle that will hit the jet at the furthest possible point! It is possible to aim a little higher, and get a little further (thanks Sammy Gerbil for pointing this out).

The question showed an equation for the trajectory that not everyone may be familiar with. I wasn't... So I decided to convince myself by deriving it from the parametric equation that I know. For a projectile fired with velocity $v$ at an angle $\theta$ to the horizontal, the horizontal and vertical components of velocity are:

$$v_x = v\cos\theta\\ vy=v\sin\theta-g\cdot t$$

We can integrate these w.r.t. time to get the position:

$$x = v\cos\theta\cdot t\\ y = v \sin \theta \cdot t - \frac12 g t^2$$

Rearranging the expression for $x$ gives us $t$:

$$t = \frac{x}{v\cos\theta}$$

Substituting this into the expression for $y$ we get

$$y = \frac{v\sin\theta\cdot x}{v\cos\theta} - \frac12 g \frac{x^2}{v^2 \cos^2\theta}\\ = x \tan\theta -\frac{g x^2}{2 v^2}\left(1+\tan^2\theta\right)$$

This is familiar to you - it was new to me. But it was the next bit where you were having trouble. You calculated the range of times at which a bullet might reach the height $h$, and assumed that the longest time taken would correspond to the furthest possible shot (which is actually not true: the time taken would be longest if the bullet is fired straight up). You also assumed that the bullet would intersect the jet that passed overhead when it was fired. Since the jet is flying faster than the bullet, this can never be true. The fact that your solution even got close to the right numerical value, then, is something of a miracle. [ but see below... I think after your latest edit I understand this "miracle" ]

So let's continue with the reasoning. We need to find the furthest possible distance $x$ that can be reached at height $h$. The only variable is $\theta$. This means that if we take the derivative of $x$ with respect to $\theta$, the stationary point (zero slope) corresponds to the furthest distance (you have to check that it's not the closest distance, of course). This is what the answer in the book did. But (and this is the clever bit), recognizing that $\tan\theta$ is a monotonic function over the range of values of interest $<-\pi/2,\pi/2]>$, one can equally decide to take a derivative with respect to $\tan\theta$. Usually this would be done formally by substitution of variables; the book's solution takes a short cut.

If we put $q = \tan\theta$, we can rewrite the equation of the trajectory as

$$y = x q - \frac{gx^2}{2 v^2}(1+q^2)\tag1$$

and the value of $q$ where the stationary point is (furthest reach) happens when


Note - using this approach, we are allowing $y$ to vary with $\theta$ - we find the highest possible value of $y$ at a given $x$. This is mathematically easier to do because the expression is linear in $y$. However, this gets us to an expression for the dashed red line in my diagram just as if I had taken the derivative of $x$ with respect to $\theta$ - but it would have been harder work.

This means

$$0 = x - \frac{q g x^2}{v^2}\\ x = \frac{q g x^2}{v^2}\\ q = \frac{v^2}{g x}$$

Substituting this expression for $q$ back into the trajectory (1), we get

$$y = \frac{v^2}{g} - \frac{g x^2}{2 v^2}\left(1 + \left(\frac{v^2}{g x}\right)^2\right)\\ = \frac{v^2}{2g} - \frac{g x^2}{2 v^2}$$

This describes a parabola that is the envelope of all possible points that can be reached (because at each value of $y$ it gives us the largest and smallest $x$ that can be hit). I added that curve as a red dashed line in the figure. And now the solution is simple.

We just have to find the range of values $x$ that are inside the red dashed line - in other words, we solve

$$h = \frac{v^2}{2g} - \frac{g x^2}{2 v^2}\\ x^2 = \frac{2 v^2}{g}\left(\frac{v^2}{2g}-h\right)$$

This gives us two values for $x$ - a positive and a negative one. If you set $g=10 ~\rm{m/s}$, the interval is $[-500\sqrt{2}, 500\sqrt{2}]$ and since the jet is flying at a velocity of 500 m/s, it is vulnerable for a total of $\sqrt{8}$ seconds - the time it takes to fly through the zone where the cannon could reach it.

Note that the cannon will at all times have to be fired considerably before the jet enters the "protected airspace" - even if fired straight up, it will take over 2.5 seconds to rach the height of the jet.

For your reference, here is the Python code used to generate the plot:

import numpy as np
import matplotlib.pyplot as plt
from math import pi,sqrt,acos,asin

vgun = 100.0
h    = 250.0
g    = 9.81

t = np.linspace(0,20,200)

# the curve that just touches the path of the jet:
vy_critical = sqrt(2*g*h)
th_critical = acos(vy_critical / vgun)
vx = vgun*np.sin(th_critical)
vy = vgun*np.cos(th_critical)
x = t*vx
y = vy*t-0.5*g*t*t
x = -x;

# add a number of trajectories at different angles:
for theta in np.linspace(-90,90,19)*pi/180:
    vx = vgun*np.sin(theta)
    vy = vgun*np.cos(theta)
    x = t*vx
    y = vy*t-0.5*g*t*t

# limiting angle - largest range
th_crit = asin(sqrt((-2*g*h+vgun*vgun)/(-2*g*h+2*vgun*vgun)))
vxc = vgun*np.sin(th_crit)
vyc = vgun*np.cos(th_crit)
xc = t*vxc
yc = t*vyc -0.5*g*t*t

# add the path of the jet:

# limit the range of the axes:
plt.xlabel('horizontal position (m)')
plt.ylabel('vertical position (m)')
plt.title('possible trajectories')

# add the envelope of trajectories
x = np.linspace(-800,800,1000)
y = vgun*vgun/(2*g) - g*x*x/(2*vgun*vgun)
plt.plot(x,y,'r',ls='--') # red dashed line


So - what, if anything, is wrong with the latest version of your solution (which is getting the right answer, after all)? I had to think about it for quite a while, but I think I figured out what is going on. My confusion was not helped by the fact that you used $x$ to mean two different things - both the initial position of the jet when the cannon is fired, and the position where it is intercepted by the cannon.

But this is why your solution works. You are solving for all possible trajectories that intersect, and for which $\tan\theta$ is valid. This leads to an inequality in your term $p^2$, and that gives you a minimum and maximum value of $p$. Finally, since $p = 500 t + x_0$, and you have both $p_{max}$ and $p_{min}$, you can find the difference in time without needing to know what $x_0$ was.

Circuitous, but correct.

The usual approach is to find the maximum value of $x$, by taking the derivative. That's what both the book and my solution did. But you did in fact find another valid way to get the solution. So well done.

  • $\begingroup$ So my solution is correct the only problem is that I didn't consider the negative time interval for firing the cannon, and also I thin(correct me if I am wrong) what you suggest is that that I should not have taken $t=0$ as the reference for firing the cannon instead I should have just considered $t$ as the time interval in which the bullet reaches the $250m$ height and then equate the $x$ coordinate of the bullet to that of the position of the jet after passage of the time $t$. $\endgroup$
    – user350331
    Jul 29 '16 at 18:14
  • 1
    $\begingroup$ Yes I think that's right. You were very close - I just wanted to visualize the situation in a way that made it clearer that you can 'shoot backwards' $\endgroup$
    – Floris
    Jul 29 '16 at 18:15
  • 2
    $\begingroup$ The trouble is that you are arbitrarily deciding there is a time you call "t=0". There is no need for that. Instead you need to see "during what segment of its flight can the jet be hit by the cannon", and that's what my diagram shows. You then say "the distance is X, velocity v, so danger time is t = X/v". There is no need to invoke "negative time" unless you say "above the cannon is t=0" but that is completely arbitrary. If the plane crosses that point at 10 o'clock you could equally say "t=36000". $\endgroup$
    – Floris
    Jul 29 '16 at 18:28
  • 1
    $\begingroup$ That was an excellent answer, Floris. That is why I upvoted it. $\endgroup$ Jul 29 '16 at 18:57
  • 1
    $\begingroup$ Please see if my latest edits clear things up for you! $\endgroup$
    – Floris
    Jul 31 '16 at 22:05

Projectile's path is a parabola, whose apex (point of maximum height) depends on $\theta$. For particular values of $\theta$, apex does not even reach up to the altitude at which enemy-jet is flying. So for all those cases where apex is equal to or greater than the altitude of enemy-jet, there is a $\textit{possibility}$ of hitting the enemy-jet (the gunner will have to time his shot properly). This is the danger-zone for the enemy-jet, beginning and ending at those values of $x$, where apex of the parabola just equals jet's altitude.

In other words whenever height of parabola's apex $y_{max}\geq y_{jet}$ (jet's altitude) the jet is in danger zone. That is what the textbook solution is saying, which is correct.

  • $\begingroup$ I see that what you imply is that that whenever $y_{max}\ge y_{jet}$, then the jet is in danger of being hit, but what if by the time the bullet rises up(or falls down from the apex) to the altitude at which the jet is flying the jet has already escaped its(the bullet's) range(at the altitude of $25m$) $\endgroup$
    – user350331
    Jul 29 '16 at 8:24
  • $\begingroup$ In the question, and going by the answer as well, danger zone seems to be defined as that region where there is mere possibility of getting hit. If you were pilot of that jet, wouldn't you define it the same way? $\endgroup$
    – Deep
    Jul 29 '16 at 10:11
  • $\begingroup$ @user350331 , are you sure that there is not a typo on the original question. As stated, the bullet is traveling 5X slower than the jet. Obviously, this is quite unrealistic if you are serious about shooting the jet down. $\endgroup$ Jul 29 '16 at 16:35

In your solution consider what happens when $t=0$.
The projectile is at position $(0,0)$ and the jet is at position $(0,250)$ ie at a height of 250 m directly above the projectile.
If this was the initially condition then the projectile would never hit the jet.

I have changed the annotation of the graph which I hope will help you understand how the solution was obtained.

That solution finds the maximum height $y_{\text{max}}$ for a given horizontal displacement of the projectile $x$.

$$y_{\text{max}}= \dfrac{u^2}{2g}-\dfrac{gx^2}{2u^2} = 500 - \dfrac{x^2}{2000}$$

enter image description here

All that now needs to be done is to make sure that the maximum height of the projectile $y_{\text{max}}$ is equal to or exceeds the height of the jet which is 250 m.
The limiting values of $x$ occur when $250 = 500 - \dfrac{x^2}{2000} \Rightarrow x = \pm 500 \sqrt 2$ m and then the answer follows.

  • $\begingroup$ From your solution what I could gather was that first of all the danger zone is the area in which the jet plane will probably be hit if the shot is timed perfectly. Also in your solution you told that the danger zone is decided in such a way that the apex of the trajectory of bullet $\ge 250m$. So, if we find out the maximum range of the projectile which comes out to be $1000~\mathrm{m}$ at $\theta=\pi/4$ and at this angle the apex comes out to be at $250m$, then $\pi/4\le\theta\le\pi/2$, am I right. $\endgroup$
    – user350331
    Jul 29 '16 at 12:18
  • $\begingroup$ Also would I have to consider the gun firing in both the quadrants. And if the gun can fire at an inclination $\pi/4\le\theta\le\pi/2$ then I am having some trouble finding the coordinates for when the bullet is at an altitude $250m$ so that I can find the interval of $x$ in which the jet is in danger. I don't know why but I am still a bit uncomfortable with this question, and am not able to think up of words that would state my doubts clearly. $\endgroup$
    – user350331
    Jul 29 '16 at 12:27
  • $\begingroup$ @user350331 I have rewritten my answer to try and explain in a little more detail as to how the book answer was obtained. $\endgroup$
    – Farcher
    Jul 29 '16 at 16:29
  • $\begingroup$ Thanks a lot....the visualization with the added annotations did help a lot and they also answered the questions which I wanted to ask but was not able to come up with good enough words to make sense out of them. I would have to confess you are a life saver. $\endgroup$
    – user350331
    Jul 29 '16 at 18:45

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