Question

Show that the maximum height and range of a projectile are $\frac{u^{2}si{n}^2\theta }{2g}$ and $\frac{u^{2}sin2\theta }{g}$ respectively

Answer 1

$\begin{array}{c}ux=u\cos \theta ax=0\\ uy=u\sin \theta ay=-g\\ \\ atmax.height\\ Vg=0\\ \Rightarrow 0-u\sin \theta =gt\\ \Rightarrow t=\frac{u\sin \theta }{g}\\ \therefore H=ugt-\frac{1}{2}ay{t}^2\\ \Rightarrow H=\frac{u\sin \theta }{g}.u\sin \theta -\frac{1}{2}\times g\times \frac{u^2{\sin }^2\theta }{g^2}\\ \therefore H=\frac{u^2{\sin }^2\theta }{2g}\\ Whenballcomestogoundt=\frac{du\sin \theta }{g}\\ \therefore Range=u_x.T+\frac{1}{2}a\times T^2\\ =u\cos \theta \times \frac{2u\sin \theta }{g}\\ \therefore Range=\frac{u^2{\sin }^2\theta }{g}\end{array}$