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Question: A particle is projected with a velocity of 20 m/s at an angle of 30<sup>0</sup> to an inclined plane...

A particle is projected with a velocity of 20 m/s at an angle of 300 to an inclined plane of inclination 30º to the horizontal. The particle hits the inclined plane at an angle 300, during its journey. The time of flight is –

A

20sin⁡600g\frac{20\sin 60^{0}}{g}

B

20sin⁡600gcos⁡300\frac{20\sin 60^{0}}{g\cos 30^{0}}

C

20sin⁡300gcos⁡600\frac{20\sin 30^{0}}{g\cos 60^{0}}

D

20sin⁡300g\frac{20\sin 30^{0}}{g}

Answer

20sin⁡600g\frac{20\sin 60^{0}}{g}

Explanation

Solution

Q Particle hits the inclined plane at an angle of 300, which means that the particle hits the plane horizontally. Thus this is the highest point of trajectory.

T = usin⁡θg\frac { u \sin \theta } { g } = 20sin⁡60∘g\frac { 20 \sin 60 ^ { \circ } } { g }