Weekly Challenge 26: Max throw

7058 /www/nrich/html/content/id/7058/ 1 2011-03-15T09:03:33 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/5987&part=">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/5837&part=">Try this next</a> </li> <li> <a href="https://nrich.maths.org/discus/messages/27/27.html">Ask NRICH</a> </li> <li> <a href="http://nrich.maths.org/6631&part=">Read all about it</a> </li> <li> <a href="http://nrich.maths.org/7059&part=solution">Last week's solution</a> </li> </ul> <div> A particle is projected with speed $10$ m $s^{-1}$  from a flat horizontal surface. Find, with proof, the angle from which it should be projected to maximise the distance travelled before it hits the surface. Does this angle depend on the speed of projection? The particle is now projected with speed $10$ m $s^{-1}$  from a height of $2$ metres. From what angle (to 3sf) should it now be projected to maximise the distance travelled before it hits the surface? Does this angle depend on the speed of projection? </div> <div class="framework"> Did you know ... ? The modelling assumptions of constant gravitational field and no friction opposing motion are good ones, leading to simple equations which always have parabolas for solution. Once these modelling assumptions are, rightly, challenged, the resulting equations become 'non-linear' and very difficult to solve. Mathematicians often take the parabola as a starting point to solving the more complicated equations and vary the solution a little to try to fit it back into the new equations. You can see an aspect of this process in the solution to this problem.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Suppose that the particle is projected from a height $H$ above the ground at speed $V$ at an angle $\alpha$ to the $x$-axis, with $x$ measuring the horizontal distance travelled and $y$ measuring the vertical distance travelled. Then the coordinates of the points along this trajectory are $$ (x, y) = (V \cos(\alpha) t, -0.5 g t^2+V\sin(\alpha) t +H) $$ The particle intersects the $x$-axis when the $y$-coordinate is zero. This is when $$ t = \frac{V\sin(\alpha) \pm \sqrt{V^2\sin^2(\alpha) +2gH}}{g} $$ The particular case $H=0$ The square root simplifies giving the point of intersection as $$ (x, y) = \left(\frac{2V^2}{g}\cos(\alpha)\sin(\alpha), 0\right) = \left(\frac{2V^2}{g}\sin(2\alpha), 0\right)\,, $$ where the second equality makes use of a trig identity. The $x$-value $V\sin(2\alpha)$ is maximised when $2\alpha=90^\circ$. Thus the optimal angle of projection is $45^\circ$, for any initial velocity. When $H\neq 0$ In this case, the particle intersects the $x$-axis at $$ x =\frac{V^2}{g} \cos(\alpha) \left(\sin(\alpha) + \sqrt{\sin^2(\alpha) +\frac{2gH}{V^2}}\right) $$ This looks complicated to differentiate so I tried a numerical solution using a spreadsheet.This produced an optimal angle of $40.4^\circ$ (3sf). It seems clear that this angle will be dependent on the initial speed, but to check I calculate the optimum angle numerically for a large initial speed of $100$m$s^{-1}$. This produces an optimium of $44.9^\circ$ (3sf) which is close to $45^\circ$, as we might intuitively expect. Extension: You might like to investigate this further. Here are some starting points. Note that the expression for the derivative is: $$ \frac{g}{V^2}\frac{dx}{d\alpha} = \cos(2\alpha)-\sin(\alpha)\sqrt{\sin^2(\alpha) +\frac{2gH}{V^2}}+\cos^2\alpha\sin\alpha \frac{1}{\sqrt{\sin^2(\alpha) +\frac{2gH}{V^2}}} $$ For an optimum we can set the left hand side to zero. This gives us $$ \sin\alpha\left(\cos(2\alpha) -\frac{2gH}{V}\right) + \sin(\alpha)\cos(2\alpha)\left(1 +\frac{2gH}{V^2\sin^2(\alpha)}\right)^{\frac{1}{2}}=0 $$ If $X(\alpha) \equiv \frac{2gH}{V^2\sin^2(\alpha)}$ is small then we can expand to give $$ \sin\alpha\left(\cos(2\alpha) -\frac{2gH}{V}\right)+ \sin(\alpha)\cos(2\alpha)\left(1+\frac{1}{2}\frac{2gH}{V^2\sin^2(\alpha)}+\mathcal{O}(X^2)\right)=0 $$ In principle that can now be turned into a polynomial expression in $\cos(\alpha)$..... </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Suppose that the particle is projected from a height $H$ above the ground at speed $V$ at an angle $\alpha$ to the $x$-axis, with $x$ measuring the horizontal distance travelled and $y$ measuring the vertical distance travelled. Then the coordinates of the points along this trajectory are $$ (x, y) = (V \cos(\alpha) t, -0.5 g t^2+V\sin(\alpha) t +H) $$ The particle intersects the $x$-axis when the $y$-coordinate is zero. This is when $$ t = \frac{V\sin(\alpha) \pm \sqrt{V^2\sin^2(\alpha) +2gH}}{g} $$ The particular case $H=0$ The square root simplifies giving the point of intersection as $$ (x, y) = \left(\frac{2V^2}{g}\cos(\alpha)\sin(\alpha), 0\right) = \left(\frac{2V^2}{g}\sin(2\alpha), 0\right)\,, $$ where the second equality makes use of a trig identity. The $x$-value $V\sin(2\alpha)$ is maximised when $2\alpha=90^\circ$. Thus the optimal angle of projection is $45^\circ$, for any initial velocity. When $H\neq 0$ In this case, the particle intersects the $x$-axis at $$ x =\frac{V^2}{g} \cos(\alpha) \left(\sin(\alpha) + \sqrt{\sin^2(\alpha) +\frac{2gH}{V^2}}\right) $$ This looks complicated to differentiate so I tried a numerical solution using a spreadsheet.This produced an optimal angle of $40.4^\circ$ (3sf). It seems clear that this angle will be dependent on the initial speed, but to check I calculate the optimum angle numerically for a large initial speed of $100$m$s^{-1}$. This produces an optimium of $44.9^\circ$ (3sf) which is close to $45^\circ$, as we might intuitively expect. Extension: You might like to investigate this further. Here are some starting points. Note that the expression for the derivative is: $$ \frac{g}{V^2}\frac{dx}{d\alpha} = \cos(2\alpha)-\sin(\alpha)\sqrt{\sin^2(\alpha) +\frac{2gH}{V^2}}+\cos^2\alpha\sin\alpha \frac{1}{\sqrt{\sin^2(\alpha) +\frac{2gH}{V^2}}} $$ For an optimum we can set the left hand side to zero. This gives us $$ \sin\alpha\left(\cos(2\alpha) -\frac{2gH}{V}\right) + \sin(\alpha)\cos(2\alpha)\left(1 +\frac{2gH}{V^2\sin^2(\alpha)}\right)^{\frac{1}{2}}=0 $$ If $X(\alpha) \equiv \frac{2gH}{V^2\sin^2(\alpha)}$ is small then we can expand to give $$ \sin\alpha\left(\cos(2\alpha) -\frac{2gH}{V}\right)+ \sin(\alpha)\cos(2\alpha)\left(1+\frac{1}{2}\frac{2gH}{V^2\sin^2(\alpha)}+\mathcal{O}(X^2)\right)=0 $$ In principle that can now be turned into a polynomial expression in $\cos(\alpha)$..... </mdoxml> 2 4 0 0 0 0 1 Weekly Challenge 26: Max throw A weekly challenge: these are shorter problems aimed at Post-16 students or enthusiastic younger students. Collections Weekly Challenge Stage 5 Mechanics mapping document Kinematics M2