A particle is projected at an angle $60°$ with speed $10\sqrt{3} \mathrm{m}/\mathrm{s}$, from the point A, as shown in the figure. At the same time, the wedge is made to move with speed $10\sqrt{3} \mathrm{m}/\mathrm{s}$ towards right as shown in the figure. Then, the time after which particle will strike with wedge is

Let u = Velocity of particle 

v = Velocity of wedge 

Location of the particle after time t

$$\begin{gathered} x=u_{x}t=\left(10\sqrt{3}\times \frac{1}{2}\right)t=5\sqrt{3}t \\ y=u_{y}t-\frac{1}{2}g{t}^2=\left(10\sqrt{3}\times \frac{\sqrt{3}}{2}\right)t-\left(\frac{1}{2}\times 10\right)t^2=15t-5t^2 \end{gathered}$$

Equation of the top surface of the wedge at time t

$$\begin{gathered} y=-\tan 30°(x-vt)=-\frac{x-10\sqrt{3}t}{\sqrt{3}} \\ \text{or, }y+\frac{x}{\sqrt{3}}=10t \end{gathered}$$

When the particle strikes the wedge, the locations of the particle satisfies the previous equation.

$$\begin{gathered} \therefore \left(15t-5t^2\right)+\frac{5\sqrt{3}t}{\sqrt{3}}=10t \\ \text{or, }{t}^2-2t=0 \end{gathered}$$

The non-zero root of the equation,  $t=2 \mathrm{s}$  is the time when the particle strikes the wedge.