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A particle is projected at an angle of elevation sin-1 (4/5) and its range on the horizontal plane is 3km.The initial velocity of projection is _______.the answer is 175 m/s.explain how? |
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Answer» tal range , R = u²sin2θ/gwhere θ is ANGLE of elevation of projectile, u is velocity of projection and g is acceleration due to gravity.here, R = 3km = 3000m , g = 9.8 m/s² θ = sin^-1(4/5) we know, sin2θ = 2sinθcosθ = 2 × sin{sin^-1(4/5)} × COS{sin^-1(4/5)}= 2 × 4/5 × cos{cos^-1(3/5)}= 2 × 4/5 × 3/5 = 24/25 so, sin2θ = 24/25then, 3000 = u²(24/25)/9.8or, 3000 × 9.8 × 25/24 = u²or, 1000 × 25/8 = u² or, 125 × 9.8 × 25 = u² or, u² = (25 × 25 × 49) or, u = 175 m/s hence, velocity of projection is 175m/s |
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