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An anti-aircraft gun fires at an elevation of 60 degree at an enemy aircraft at 10000m above the ground. At what speed must the cannon be shot to hit the plane at that height? (g=9.8)

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5 answers
  1. 1/2 m (v sin(60º)^2 = m g 10 km

    v = √(g * 20 km) / sin(60º)

    this is the minimum velocity for the projectile to reach the height of the plane ... faster would be better

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  2. Good

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  3. An anti air craft gun fires at an elevation of 60 degree at an enemy air craft at 1000m above the plane at the height (g=9.8ms2

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  4. an anti aircraft fires at an elevation of 60degrees at an enemy aircraft at 10000m above d ground. at what speed must the cannon be shot to hit the plane at that height. take g as 9.8m/s

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  5. I haven’t seen any solution

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