Projectile motion help (1 Viewer)

tsoliman1 · Jun 5, 2016

Hey guys

So I don't know how to do this and I need help.

A gun can fire shells with speed "u" at elevations between 30 and 60 degrees. Find the area commanded by the gun on a horizontal plane, assuming that the gun can be rotated about vertical axis so as to point in any desired direction.

Leave answer in terms of g.

The answer is pi(u^4)/4g^2

I need the working out please.

Thanks

InteGrand · Jun 5, 2016

tsoliman1 said:
Hey guys

So I don't know how to do this and I need help.

A gun can fire shells with speed "u" at elevations between 30 and 60 degrees. Find the area commanded by the gun on a horizontal plane, assuming that the gun can be rotated about vertical axis so as to point in any desired direction.

Leave answer in terms of g.

The answer is pi(u^4)/4g^2

I need the working out please.

Thanks

$\noindent The range of a projectile is $d=\frac{u^2}{g}\sin 2\theta$, where $\theta$ is the angle of inclination, $u$ is the speed of projection, and we have assumed the projectile starts on ground level (and no air resistance of course). You can see this formula here:$

https://en.wikipedia.org/wiki/Range_of_a_projectile .

$\noindent (In the exam you'd have to derive this result.) Now, the maximal range is when $\theta = 45^\circ \Rightarrow \sin 2\theta = 1$, which is $d_{\text{max.}}=\frac{u^2}{g}$. The minimum range is when $\theta=30^\circ$ or $60^\circ$ (note that angles of $\theta$ and $90^\circ-\theta$ result in equal range). At this angle, $\sin 2\theta = \sin 60^\circ = \frac{\sqrt{3}}{2}$. So the minimum range is $d_{\text{min.}}=\frac{u^2}{g}\cdot \frac{\sqrt{3}}{2}$. As $\theta$ varies in between, all range values between $d_{\text{min.}}$ and $d_{\text{max.}}$ get taken. So as the gun sweeps around, the region covered is an annulus. Its area is, using the area of an annulus formula,$

$\begin{align*} A&=\pi \left(d_{\text{max.}}^2 - d_{\text{min.}}^2 \right) \\ &=\pi \left(\left(\frac{u^2}{g}\right)^2 -\left(\frac{u^2}{g}\cdot \frac{\sqrt{3}}{2}\right)^2 \right)\\ &=\pi \left(\frac{u^4}{g^2}-\frac{3}{4}\times \frac{u^4}{g^2}\right)\\ &= \pi \cdot \frac{u^4}{4g^2}.\end{align*}$

tsoliman1 · Jun 5, 2016

Sorry I don't understand that. We haven't learnt about annulus yet. Also I don't understand how it got minimum range to be 30 or 60 degrees.

Whoops I just realised what annulus is sorry. But still don't get the minimum range bit.

InteGrand · Jun 5, 2016

tsoliman1 said:
Sorry I don't understand that. We haven't learnt about annulus yet. Also I don't understand how it got minimum range to be 30 or 60 degrees.

$\noindent Intuitively, the range is smaller if the projection angle is smaller (because you hit the ground pretty quickly, so not much time for the projectile to go forward much), increasing until you get to the angle with maximal range (famously known to be $45^\circ$), and then it just goes down again if you launch too high.$

$\noindent Mathematically, since the range is $d\left(\theta\right) = \frac{u^2}{g}\sin 2\theta$, note that this means $d$ is an increasing function as $\theta$ varies between $0$ and $45^\circ$ (because $\sin x$ is increasing on the interval $0\leq x \leq 90^\circ$), and is then decreasing (as sine starts decreasing after its argument (input value) gets to $90^\circ$), and also $d\left(90^\circ - \theta\right) = d\left(\theta\right)$. This means the max. $d$ occurs at $45^\circ$ (as the sine is 1 there, which is the max. value for sine), and the minimum is the lowest allowable $\theta$, which is $30^\circ$.$

An annulus is just a ring shape. See this image: https://upload.wikimedia.org/wikipe.../Annulus_area.svg/2000px-Annulus_area.svg.png .

The reason the region swept out by the gun is an annulus is that at the minimum range, we get the inner circle of the annulus as we sweep the gun around at the minimum range. As we keep increasing the range and swing the gun around, we keep getting concentric circles (circles because we're sweeping the gun around at a given range), and we stop when we get to the furthest possible range, which'll correspond to the outer circle of the annulus.

The area of the annulus is basically just subtracting the area of the inner circle (disk) from the outer one.

tsoliman1 · Jun 5, 2016

I get it now! Thanks so much!!

eyeseeyou · Jun 6, 2016

Is projectile motion hard?

tsoliman1 · Jul 25, 2016

eyeseeyou said:
Is projectile motion hard?

No not really

Projectile motion help (1 Viewer)

tsoliman1

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