A cannon is fired with a muzzle velocity of 300 m/s at an angle of 60 degree. What is its range?

Question

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Answer 1

For a projectile travelling through air without air resistance. The 'x' direction will represent horizontal distance and the 'y' direction will represent vertical distance.

The time 't' of flight is given by: -

t = 2.v.sin(θ)

....__

.... g

Where 'v' is the initial velocity and 'θ' is the angle of release, with 'g' as the acceleration due to gravity (9.81 m/²).

The maximum range is given by: -

x = v².sin(2.θ)

....__

.... g

The greatest height attained by the projectile is: -

y = v².sin²(θ)

....__

.... 2g

And this height maximum occurs at time: -

t = v.sin(θ)

....__

.... g

Hence, the projectile follows a parabolic flight path!

calculation.

_________

If v = 300 m/s and θ = 60 degrees.

The the range 'x' from ther above equations is: -

x = (300)².sin(2 x 60)

......___

......9.81

Thus,

The range is x = 7945.187 m (rounded to 3DP)

When air resistance is taken into consideration, this calculation is more difficult and generally only yields approximate results.