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

Answer 1

m g h < 1/2 m [v sin(60º)]^2

v sin(60º) > √(2 g h)

Answer 2

Range = Vo^2*sin(2A)/g = 10,000.
Vo^2*sin(120)/9.8 = 10,000,
Vo^2*0.0884 = 10,000,
Vo = 336.4 m/s.

Answer 3

correct