Question

A particle is projected horizontally with a speed u from the top of a plane inclined at an angle $\theta$ with the horizontal. How far from the point of projection will the particle strike the plane?

• A. $\frac{2u^2}{g}\tan \theta sec\theta$
• B. $\frac{2u}{8}{\tan }^2\theta sec\theta$
• C. $\frac{2u^2}{g}\tan \theta \cos \theta$
• D. $\frac{2u}{g}\tan \theta {\cos }^2\theta$

Answer 1

The correct option is A $\frac{2u^2}{g}\tan \theta sec\theta$

The horizontal distance covered by the projectile in time t is ut

The vertical distance covered during that time is $g\frac{t^2}{2}$

$tan\theta$ = $\frac{verticaldistance}{horizontaldistance}$ = $\frac{g\frac{t^2}{2}}{ut}=\frac{gt}{2u}$

$t=\frac{2utan\theta }{g}$

Horizontal distance = $\frac{2u^{2}tan\theta }{g}$ , vertical = $\frac{2u^{2}ta{n}^2\theta }{g}$

So distance between point and point of projection = $\sqrt{(\frac{2u^{2}tan\theta }{g}{)}^2+(\frac{2u^{2}ta{n}^2\theta }{g}{)}^2}$

$=\frac{2u^{2}tan\theta sec\theta }{g}$