A particle has an initial velocity of 100 m/s up to the right at 30° with the horizontal. The components of the acceleration are constant at ax=-4m/s^2 and ay=-20m/s^2. Computing the horizontal distance covered until the particle reaches a point 60m below its original elevation would be

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Sahithya · Answer

Vo=100m/s Vox=100cos30°=86.6m/s Voy=100sin30°=50m/s ay=-20m/s^2 Sy=voyt+1/2ayt^2 -60=50t-1/2(20) t^2 t=6s ax=-4m/s^2 Sx=voxt+1/2axt^2 Sx=86.6(6) +1/2(-4) (6×6) Sx=448m

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A particle has an initial velocity of 100 m/s up to the right at 30° with the horizontal. The components of the acceleration are constant at ax=-4m/s^2 and ay=-20m/s^2. Computing the horizontal distance covered until the particle reaches a point 60m below its original elevation would be

A particle has an initial velocity of 100 m/s up to the right at 30° with the horizontal. The components of the acceleration are constant at ax=-4m/s^2 and ay=-20m/s^2. Computing the horizontal distance covered until the particle reaches a point 60m below its original elevation would be

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A particle has an initial velocity of 100 m/s up to the right at 30° with the horizontal. The components of the acceleration are constant at ax=-4m/s^2 and ay=-20m/s^2. Computing the horizontal distance covered until the particle reaches a point 60m below its original elevation would be

A particle has an initial velocity of 100 m/s up to the right at 30° with the horizontal. The components of the acceleration are constant at ax=-4m/s^2 and ay=-20m/s^2. Computing the horizontal distance covered until the particle reaches a point 60m below its original elevation would be

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