projectile question (1 Viewer)

trecex1 · Jul 17, 2016

Screen Shot 2016-07-17 at 7.40.01 pm.png

Just need help with part iv) (don't really know what question means) you can assume all other parts are true.

InteGrand · Jul 17, 2016

trecex1 said:
View attachment 33364
Just need help with part iv) (don't really know what question means) you can assume all other parts are true.

$\noindent Basically it's saying to show that no point above the ground can be hit by firing the projectile at two different smaller-than-$45^\circ$ angles. E.g. it's impossible to hit the same point with a projection angle of $30^\circ$ and also with an angle $15^\circ$ (assuming $V$ is held fixed). (In part (ii), it was shown that we can sometimes get points with two different angles. If you can show part (iii), then it means that the two different angles in part (ii) won't both be less than $45^\circ$.)$

$\noindent To do this Q. (iv), you can assume that $\theta_1,\theta_2<\frac{\pi}{4}$ and that an arbitrary but specified point above the ground is hit both when $\theta = \theta_1$ and when $\theta = \theta_2$. Then try and whittle it down to the fact that $\theta_1 = \theta_2$, which means we can't do it with two such \emph{different} less-than-$45^\circ$ angles.$

trecex1 · Jul 17, 2016

Ok, not sure if solution is basically your method or not, they just moved everything in i) to one side assumed any point (x,y) with 2 solutions then showed product of roots was >1, next step was one of the roots had to be greater than 1. So angle had to be greater than pi/4 for 2 solutions... How did they go from if product of 2 roots (tan(theta)) is >1 that either root has to be >1? If you could explain this or give a simpler solution would be great - seems a lot of work for 1 mark!

Screen Shot 2016-07-17 at 8.03.50 pm.png

<--- their solution

InteGrand · Jul 17, 2016

$\noindent Here's one way to do it.$

$\noindent Note that the range of the projectile is $R=\frac{V^2 \sin 2\theta}{g}$ (left as exercise for the reader, it's quite a standard thing to prove). So any point above the ground that is reachable by the projectile has $x$ value less than $R$, and greater than $0$ (a point with $x=0$ above the ground would only be reachable if the projectile were fired directly up, but this didn't happen as we are told $\theta < \frac{\pi}{2}$). If we want to consider $\theta$ varying in $\left(0,\frac{\pi}{4}\right)$, then the $x$-value of any point possible for the projectile to hit will satisfy $x< \frac{V^2}{g}$..$

$\noindent Note that by rearranging the flight equation from (i) and writing $T\equiv \tan \theta$, we have the projectile's trajectory being $y=-\frac{x^2}{4h}T^2 + xT - \frac{x^2}{4h}$. Consider $\theta$ in the range $0<\theta < \frac{\pi}{4}$ (so that $0<T<1$). Fix $X$ with $0<X < \frac{V^2}{g}$, then the $y$-value of the point on the parabola of the projectile is $Y=-\frac{X^2}{4h}T^2 + XT - \frac{X^2}{4h}$. Consider this as a function of $T$, i.e. $Y = f(T) := -\frac{X^2}{4h}T^2 + XT - \frac{X^2}{4h}$. Basically the question is equivalent to getting us to prove that $f$ is one-to-one for $0<T<1$ (can you see why?). This can be shown as follows.$

$\noindent Note that $f$ is a quadratic function of $T$. The turning point of this function is at$

$\begin{align*}T_{vert.} &= -\frac{X}{2\times -\frac{X^2}{4h}} \\ &= \frac{2h}{X} \\ &= \frac{V^2}{g} \cdot \frac{1}{X} \quad \Big{(}\text{as }h = \frac{V^2}{2g}\Big{)} \\ &> \frac{V^2}{g} \times \frac{g}{V^2} \quad \Big{(}\text{as }X< \frac{V^2}{g}\Big{)} \\ &= 1.\end{align*}$

$\noindent So $f(T)$'s graph is a parabola whose turning point occurs after $T=1$, so $f$ is certainly one-to-one for $0<T<1$ (as a quadratic is one-to-one either side of its turning point). This proves the claim. (In fact we have shown that this is an increasing function of $T$, because the turning point occurs after $T=1$, and $f(T)$'s graph is a concave down parabola. This shows that for any given $x$-value, the $y$-value of the hit-point (which may be below ground level here but this doesn't affect the proof of the claim) is higher when $T$ is higher, equivalently when $\theta$ is higher, for $\theta \in \left(0,\frac{\pi}{4}\right)$. This is intuitively obvious too, as seen for instance by looking at the lower three parabolas in the picture below.)$

http://www.animations.physics.unsw.edu.au/images/download_projectiles1.gif

InteGrand · Jul 17, 2016

trecex1 said:
Ok, not sure if solution is basically your method or not, they just moved everything in i) to one side assumed any point (x,y) with 2 solutions then showed product of roots was >1, next step was one of the roots had to be greater than 1. So angle had to be greater than pi/4 for 2 solutions... How did they go from if product of 2 roots (tan(theta)) is >1 that either root has to be >1? If you could explain this or give a simpler solution would be great - seems a lot of work for 1 mark!View attachment 33365 <--- their solution

Ah yeah, that solution is much nicer.

The product of roots is greater than 1. Since the roots are each positive numbers (as they are tan of something that is acute), at least one of them must be greater than 1 (otherwise both would be less than 1, and then their product would have been less than 1 instead of greater than 1. This is because the product of two numbers smaller than 1 is smaller than 1.).

trecex1 · Jul 17, 2016

InteGrand said:
Ah yeah, that solution is much nicer.

The product of roots is greater than 1. Since the roots are each positive numbers (as they are tan of something that is acute), at least one of them must be greater than 1 (otherwise both would be less than 1, and then their product would have been less than 1 instead of greater than 1. This is because the product of two numbers smaller than 1 is smaller than 1.).

Oh i see, thanks Integrand, would not have thought of this in an exam lol.

projectile question (1 Viewer)

trecex1

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trecex1

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