# Maximum value question

1. Sep 6, 2007

### John O' Meara

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.

2. Sep 6, 2007

### learningphysics

sin(A+B) = sinAcosB + cosAsinB.... not 2sinAcosB