Question

A particle is projected with a certain velocity so as to pass through a point at a horizontal distance R from the point of projection .Show that if t₁and t₂ are the time taken to reach this point in two possible ways in which the particle can be projected with a given velocity then t₁ t₂ = 2R/g.

Answer 1

$$\begin{gathered} \mathrm{Answer} \\ \mathrm{for}\mathrm{equal}\mathrm{range}\mathrm{R},\mathrm{the}\mathrm{particle}\mathrm{should}\mathrm{be}\mathrm{either}\mathrm{be}\mathrm{projected}\mathrm{at}\mathrm{an}\mathrm{angle}\mathrm{\theta }\mathrm{or}90-\mathrm{\theta } \\ {\mathrm{t}}_1=\frac{2\mathrm{usin\theta }}{\mathrm{g}},{\mathrm{t}}_2=\frac{2\mathrm{u}\sin (90-\mathrm{\theta })}{\mathrm{g}}=\frac{2\mathrm{ucos\theta }}{\mathrm{g}} \\ {\mathrm{t}}_1\times {\mathrm{t}}_2=\frac{2\mathrm{usin\theta }}{\mathrm{g}}\frac{2\mathrm{ucos\theta }}{\mathrm{g}}=\frac{2{\mathrm{u}}^2\sin 2\mathrm{\theta }}{{\mathrm{g}}^2}=\frac{2\mathrm{R}}{\mathrm{g}} \\ \mathrm{where}\mathrm{R}=\frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}} \end{gathered}$$