# Minimum distance between bomb and shell

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1. Aug 2, 2016

### Akhilesh Prasad

1. Prolem Statement:-
An aircraft is flying horizontally with constant velocity = $200 m/s$, at a height = $1 km$ above the ground. At the moment shown, a bomb is released from aircraft and the canon- gun below fires a shell with initial speed = $200 m/s$, at some angle #\theta#.
1)For what value of $'\theta'$ will the projectile shell destroy the bomb in mid- air ? If the value of $'\theta'$ is $53^{\circ}$.
2)Find the minimum distance between the bomb and the shell as they fly past each other. Take $\sin53^{\circ}=4/5$.

The figure for the question is as shown below:

2. My solution:-

I have solved the first part of the question the one that I am having trouble solving is the second part. Here is my work on the solution.
For reference these are the variables that I have used in my solution:-
• $\vec{s}_{s/e}$-Displacement of shell w.r.t Earth
• $\vec{s}_{b/e}$-Displacement of bomb w.r.t Earth
• $\vec{v}_{0}$-Velocity of bomb/shell w.r.t Earth
Lets not list all of them, I think you should have got the idea now. So, here goes my solution.

$\vec{s}_{s/e}=-(v_0\cos{\theta}t)\hat{i} + (v_0\sin{\theta}t-(1/2)gt^2)\hat{j}$
$\vec{s}_{b/e}=(v_0t)\hat{i}+(1-(1/2)gt^2)\hat{j}$

Now, to find the minimum distance between the bomb and the shell what I did was first find the relative displacement and then differentiate its modulus and set it as $0$ which gives us the minimum possible distance between them.

\begin{aligned} &\vec{s}_{s/b}=-(v_0\cos{\theta}+v_0)t\hat{i}+(v_0\sin\theta-1)\hat{j} \\ \implies & |\vec{s}_{s/b}|^2=2v_0^2t^2(1+\cos\theta)-2v_0\sin\theta t \qquad\qquad\qquad\qquad\qquad \ldots(1) \end{aligned}

Now, differentiating equation $(1)$ w.r.t $t$ and putting $\dfrac{d|\vec{s}_{s/b}|^2}{dt}=0$, we get

$t=\dfrac{\sin\theta}{2v_0(1+\cos\theta)}=\dfrac{1}{4v_0}$

On substituting $t=1/(4v_0)$ in equation $(1)$, we get

${|\vec{s}_{s/b}|}^2=\dfrac{320}{16\times 25}=\dfrac{4}{5} \implies |\vec{s}_{s/b}|=\dfrac{2}{\sqrt5}$

3.Book's Solution:-

Why does my answer differ from the book, what did I miss.

Last edited: Aug 2, 2016
2. Aug 2, 2016

### Staff: Mentor

An image showing the initial relative positions of the aircraft and gun would be helpful. Your equation for the trajectory of the shell indicates that its x-component is in the direction of the negative x-axis, while that of the bomb is positive. If they both start at the same x-coordinate then surely they can't intersect as they would be moving in opposite directions away from the origin. So there must be some initial separation in the x-direction.

3. Aug 2, 2016

### Akhilesh Prasad

I have been trying to upload for quite some time but don't know why it doesn't get uploaded.

4. Aug 2, 2016

### Akhilesh Prasad

Finally, it got uploaded.

5. Aug 2, 2016

### Staff: Mentor

So according to your diagram the initial x-separation is $\sqrt{3}~km$. You'll have to work that into your position equations.

6. Aug 2, 2016

### Akhilesh Prasad

Oh my god.....that was a blunder, thanks for all your efforts.

Last edited: Aug 2, 2016
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