​A particle is projected from ground at an angle 60 with horizontal at speed of 10root3

Dear Student,

Please find below the solution to the asked query:

Let after t time the particle hits the wedge. Then the horizontal distance moved by the wedge more than the particle is,

$$\begin{gathered} x=x_{wed}-x_{par}=10\sqrt{3}t-10\sqrt{3}\cos {60}^{0}t=10\sqrt{3}\left(1-\frac{1}{2}\right)t=5\sqrt{3}t \\ So, the vertical height of the point on the wedge, where the particle hits is, \\ y=x \tan {30}^0=5\sqrt{3}t\times \frac{1}{\sqrt{3}}=5t \\ Now, according to Kinematic equation, \\ y=10\sqrt{3} \sin {60}^{0}t-\frac{1}{2}g{t}^2=15t-5t^2 \\ Therefore, \\ 5t=15t-5t^2\Rightarrow 5t^2=10t\Rightarrow t=2 s \end{gathered}$$

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