- #1
John O' Meara
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The range of a particle from a point on a plane which is inclined at an angle alpha to the horizontal is given by [tex] R=\frac{2u^2}{g}\cos\theta\sin(\theta-\alpha)\sec^2 \alpha \\[/tex], where the velocity of projection is u at an angle theta to the horizontal. Using the trigonometrical identity 2cosAsinB=sin(A+B), find the maximum value of R as theta varies. Verify your result by differentation.(a)
[tex] R=\frac{u^2}{g}(\sin(2\theta-\alpha)-\sin\alpha)\sec^2 \alpha\\[/tex]. Maximum range for a given velocity of projection: since [tex] \sin(\pi- \theta) = \cos\theta \\[/tex]. Therefore the same values of R will be obtained whether the angle of projection is theta or pi-theta. Although the range will be the same for both angles, the time taken and height will be different. R is greatest when [tex] \sin2\theta = \pi \mbox{ therefore }\\ R_{max} = \frac{u^2}{g}(\sin\pi-\alpha \ - \ \sin\alpha)\sec^2 \alpha \\ \mbox{which } =\frac{u^2}{g}(\cos\alpha - \sin\alpha)\sec^2 \alpha\\ [/tex].
(b) I get [tex] \frac{dR}{d\theta} = \frac{2u^2\sec^2 \alpha}{g}(\cos2\theta \cos\alpha +\sin2\theta\sin\alpha) \\ [/tex]. I think my reasoning is wrong in part (a), please show me where I'm wrong. Thanks.
[tex] R=\frac{u^2}{g}(\sin(2\theta-\alpha)-\sin\alpha)\sec^2 \alpha\\[/tex]. Maximum range for a given velocity of projection: since [tex] \sin(\pi- \theta) = \cos\theta \\[/tex]. Therefore the same values of R will be obtained whether the angle of projection is theta or pi-theta. Although the range will be the same for both angles, the time taken and height will be different. R is greatest when [tex] \sin2\theta = \pi \mbox{ therefore }\\ R_{max} = \frac{u^2}{g}(\sin\pi-\alpha \ - \ \sin\alpha)\sec^2 \alpha \\ \mbox{which } =\frac{u^2}{g}(\cos\alpha - \sin\alpha)\sec^2 \alpha\\ [/tex].
(b) I get [tex] \frac{dR}{d\theta} = \frac{2u^2\sec^2 \alpha}{g}(\cos2\theta \cos\alpha +\sin2\theta\sin\alpha) \\ [/tex]. I think my reasoning is wrong in part (a), please show me where I'm wrong. Thanks.