At what other angle (other than 45 degrees) could a gun be fired to travel the same horizontal distance
Is the answer: There is not one?
That is maximum range, no other angle will reach that range.
of course all shorter ranges have a lob angle and a straight shot angle :)
To find the angle at which a gun could be fired to travel the same horizontal distance as when fired at 45 degrees, we can use the concept of projectile motion.
Let's assume the initial velocity of the projectile (bullet) is the same. When a projectile is fired at an angle of 45 degrees, it can be divided into two components: horizontal and vertical. The horizontal component remains constant throughout the motion, while the vertical component changes due to gravity.
To find another angle that will result in the same horizontal distance, we need to consider that the horizontal component remains the same. This means that the initial horizontal velocity is equal to the horizontal velocity at any other angle.
The horizontal and vertical components of the velocity can be calculated using trigonometry. At 45 degrees, the horizontal and vertical velocities are the same, as the angle is evenly divided. Therefore, both components are cos(45) times the initial velocity.
To find another angle that gives the same horizontal distance, we need to find an angle where the vertical component of the velocity cancels out the effect of gravity, resulting in the same time of flight. This can be found using the equation of motion:
Horizontal distance = horizontal velocity * time of flight
Since the horizontal velocity remains the same, the time of flight should also be the same. The time of flight can be calculated using the equation:
time of flight = (2 * vertical velocity) / g
where g is the acceleration due to gravity.
Now, if we substitute the values, we have:
time of flight @ 45 degrees = (2 * cos(45) * initial velocity) / g
To find the angle at which the same horizontal distance is covered, we want to find an angle where:
time of flight @ other angle = time of flight @ 45 degrees
Let's call the other angle θ. The vertical velocity at this angle can be calculated using the equation:
vertical velocity @ θ = sin(θ) * initial velocity
Substituting these values, we have:
time of flight @ θ = (2 * sin(θ) * initial velocity) / g
To find the angle θ that gives the same time of flight as 45 degrees, we set the two equations equal to each other:
(2 * sin(θ) * initial velocity) / g = (2 * cos(45) * initial velocity) / g
Canceling out initial velocity and g, we have:
sin(θ) = cos(45)
Taking the inverse sine of both sides:
θ = arcsin(cos(45))
Evaluating this expression, we find:
θ ≈ 44.48 degrees
So, the gun could be fired at an angle of approximately 44.48 degrees to travel the same horizontal distance as when fired at 45 degrees.
Therefore, the answer to your question is that there is another angle, approximately 44.48 degrees, at which the gun can be fired to travel the same horizontal distance.