Two balls A and B are projected simultaneously from the top of a long smooth inclined plane. The ball A is projected along the surface with speed u while ball B is thrown horizontally with velocity v $$=$$ $$10$$ $$m/s$$, from the same point. Both the balls travel in the same vertical plane and the ball B hits the ball A on the incline. If the inclination angle of the incline is θ$$=$$$$cos$$−$$1$$⁡$$45$$, find the displacement (in$$ $$meter) from the point of projection, where the two balls collide. Take $$g=10ms-2$$

Question

Two balls A and B are projected simultaneously from the top of a long smooth inclined plane. The ball A is projected along the surface with speed u while ball B is thrown horizontally with velocity v $$=$$ $$10$$ $$m/s$$, from the same point. Both the balls travel in the same vertical plane and the ball B hits the ball A on the incline. If the inclination angle of the incline is θ$$=$$$$cos$$−$$1$$⁡$$45$$, find the displacement (in$$ $$meter) from the point of projection, where the two balls collide. Take $$g=10ms-2$$

Infinity Learn · Accepted Answer

Let us fix our $$co-ordinate$$ axes along the incline (x) and perpendicular to it (y),the acceleration of both the balls in $$x-direction$$ is same equal to $$ax=gsin$$⁡θ. ∴ The two balls will collide if the $$x-component$$ of their initial velocities is equal. ⇒$$u=vcos$$⁡θ∴$$u=10$$×$$45=8m/sThe$$ time of flight of the projectile can be calculated by consideration of its motion in $$y-direction.0=(vsin$$⁡θ$$)t$$−$$12(gcos$$⁡θ$$)t2$$⇒$$t=2vsin$$⁡θgcos⁡θ$$=2$$×$$10$$×$$310$$×$$4=32sec$$∴ Distance $$PQ=ut+12(gsin$$⁡θ$$)t2$$   $$=8$$×$$32+12$$×$$10$$×$$35$$×$$322=18.75m$$

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