- #1
John O' Meara
- 325
- 0
The range of a particle from a point on a plane which is inclined at an angle alpha to the horizontal is given by R=\frac{2u^2}{g}\cos\theta\sin(\theta-\alpha)\sec^2 \alpha \\, 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)
R=\frac{u^2}{g}(\sin(2\theta-\alpha)-\sin\alpha)\sec^2 \alpha\\. Maximum range for a given velocity of projection: since \sin(\pi- \theta) = \cos\theta \\. 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 \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\\.
(b) I get \frac{dR}{d\theta} = \frac{2u^2\sec^2 \alpha}{g}(\cos2\theta \cos\alpha +\sin2\theta\sin\alpha) \\. I think my reasoning is wrong in part (a), please show me where I'm wrong. Thanks.
R=\frac{u^2}{g}(\sin(2\theta-\alpha)-\sin\alpha)\sec^2 \alpha\\. Maximum range for a given velocity of projection: since \sin(\pi- \theta) = \cos\theta \\. 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 \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\\.
(b) I get \frac{dR}{d\theta} = \frac{2u^2\sec^2 \alpha}{g}(\cos2\theta \cos\alpha +\sin2\theta\sin\alpha) \\. I think my reasoning is wrong in part (a), please show me where I'm wrong. Thanks.
