Question

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

Answer 1

Let $u=$ Velocity of particle

$v=$ Velocity of wedge

Horizontal range of projectile

$R=\frac{u^2\sin 2\theta }{g}$

$R=\frac{(10\sqrt{3}{)}^2\sin 120}{g}=\frac{150\sqrt{3}}{g}$

So, $t=R/v=\frac{150\sqrt{3}}{g}\times 110\sqrt{3}=\frac{15}{10}=1.55\approx 2s$