Question

A body is projected with a speed u at an angle theta with the horizontal. The speed of the body when it is at the highest point on its trajectory isRoot over 2/5 times its speed at half the maximumheight.Find the value of theta.

Answer 1

$$\begin{gathered} themaximumheightattainedbytheprojectileisgivenbythirdeqnofmotioni.e. \\ 0=u^2{\sin }^2\theta -2ghorh=\frac{u^{2}si{n}^2\theta }{2g}. \\ nowthevelocityat\frac{h}{2}isgivenby, \\ {v}^2=u^{2}si{n}^2\theta -gh.Butatthehighestpointvelocity{V}_0=u\cos \theta \\ againgiven\frac{2}{5}\left(u^{2}si{n}^2\theta -gh\right)=u^2{\cos }^2\theta \\ puttingthevalueofhinaboveeqnweget, \\ \frac{2}{5}\left(u^{2}si{n}^2\theta -g\frac{u^{2}si{n}^2\theta }{2g}\right)=u^{2}co{s}^2\theta \\ \frac{u^{2}si{n}^2\theta }{5}=u^{2}co{s}^2\theta or{\tan }^2\theta =5or\tan \theta =\sqrt{5}or\theta ={\tan }^{-1}\left(\sqrt{5}\right). \end{gathered}$$