# a stone is projected from point a with speed u making an angle 60 degrees with the horizontal.the fixed inclined surface makes an angle 30 degrees with the horizontal.the stone lands at b after time t.then the distance ab=

Dear Student,

Please find below the solution to the asked query:

Consider the situation as shown in the figure below.

Let a particle be projected up with a speed u from an inclined plane which makes an angle α with the horizontal velocity of projection makes an angle θ with the inclined plane.

The equation for the range on an inclined plane is,

$R=\frac{2{u}^{2}}{g}\frac{\mathrm{sin}\theta \mathrm{cos}\left(\theta +\alpha \right)}{{\mathrm{cos}}^{2}\alpha}$

Now according to the given information in the question,

$\alpha ={30}^{0}and\phantom{\rule{0ex}{0ex}}angleofprojectionwithrespecttohorizontalis\phantom{\rule{0ex}{0ex}}\theta +\alpha ={60}^{0}\Rightarrow \theta ={30}^{0}$

Therefore, the range on the inclined plane. In the given question *ab* is,

$R=\frac{2{u}^{2}}{g}\frac{\mathrm{sin}\theta \mathrm{cos}\left(\theta +\alpha \right)}{{\mathrm{cos}}^{2}\alpha}=\frac{2{u}^{2}}{g}\frac{sin{30}^{0}cos{60}^{0}}{co{s}^{2}{30}^{0}}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{2{u}^{2}}{g}\frac{\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}}\right)}{{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^{2}}=\frac{2{u}^{2}}{g}\frac{\left({\displaystyle \frac{1}{4}}\right)}{\left({\displaystyle \frac{3}{4}}\right)}=\frac{2{u}^{2}}{g}\frac{1}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{2{u}^{2}}{3g}$

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