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Question

A particle A is projected from the ground with an initial velocity of 10 m/s at an angle of 60 degree with horizontal. From what height h should an another particle B be projected horizontally with velocity 5 m/s so that both the particles collide in ground at point C if both are projected simultaneously (g=10 m/s2)

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Solution

When a particle is projected with speed u=10ms-1 making an angle of 60° with the ground, thehorozontal velocity=ucos60°=10×12=5ms-1and vertical velocity up=usin60°=10×32ms-1. The velocity at highest point =0 and corresponding hight attained=h. Therefore, from third equation of motion for vertical motionis given by,v2=u2-2gh. At the heighest point v=0, initial velocity u=usinθ, therefore,0=usinθ2-2gh. i.e. 0=(10×32)2-2×10×h, where g=10ms-2.or h=100×34×2×10 or h=154 or h=3.75m.The horizontal velocity through out the projectile motion remains the same, therefore when the particle is at the height of 3.75m from ground, the time taken to reach thereis given by,0=usinθ-gt or 10×32-10t=0 or t=1.732s. When second body throne horizontly , its fall is free, therefore in 1.732s it will fall by distance'h0=12×10×32=15m. Therfore second particle must be thrown simultaneouslyfrom a height of 3.75+15m =18.75m.Note that horizontal velocity is independent of vertical velocity.

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