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,

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

Now according to the given information in the question,

$$\begin{gathered} \alpha ={30}^0 and \\ angle of projection with respect to horizontal is \\ \theta +\alpha ={60}^0\Rightarrow \theta ={30}^0 \end{gathered}$$

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

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

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