Question

# A particle is projected from point $$A$$ with velocity $$\displaystyle u\sqrt{2}$$   at angle $$45$$ to horizontal as shown in diagrame. It strikes the plane $$BC$$ at right angle. What is the velocity of particle at the time of impact?

A
3u2
B
u2
C
2u3
D
u

Solution

## The correct option is D $$\displaystyle \frac {2u}{\sqrt{3}}$$The horizontal component of the velocity remains the same=$$\sqrt{2}u cos45^{\circ}=u$$Let the vertical velocity of the particle when it strikes the plane be $$v$$ downwards. Since it hits the plane perpendicularly, the final velocity forms an angle of $$60^{\circ}$$ with the vertical.Thus $$\dfrac{u}{v}=tan60^{\circ}=\sqrt{3}$$$$\implies v=\dfrac{u}{\sqrt{3}}$$Thus the magnitude of the velocity when it hits the plane is $$\sqrt{v_x^2+v_y^2}=\sqrt{u^2+(\dfrac{u}{\sqrt{3}})^2}=\dfrac{2u}{\sqrt{3}}$$Physics

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