A fighter, traveling at 474 m/s fires a bullet perpendicular to its flight path. The muzzle velocity of the bullet is 515 m/s.
At what angle to its intended path does
the bullet actually travel? Answer between −180◦ and +180◦.
Answer in units of â—¦.
To find the angle at which the bullet travels relative to the fighter's intended path, we can use the concept of vector addition and trigonometry.
Let's assume that the fighter's intended path is along the x-axis, and the bullet's path is along the y-axis. The velocity of the fighter is given as 474 m/s, so its velocity vector can be represented as Vf = 474i, where i is the unit vector along the x-axis.
The muzzle velocity of the bullet is given as 515 m/s. Since it is fired perpendicular to the flight path, its velocity vector can be represented as Vb = 515j, where j is the unit vector along the y-axis.
To find the resulting velocity vector of the bullet, we need to add the vectors Vf and Vb. This can be done by computing their vector sum:
Vresult = Vf + Vb
Vresult = (474i) + (515j)
To determine the angle between the bullet's path and the fighter's intended path, we can use the arctangent function:
θ = arctan(Vresult.y / Vresult.x)
Here, Vresult.y represents the y-component of the resulting velocity vector and Vresult.x represents the x-component of the resulting velocity vector.
Calculating the values:
Vresult.x = 474 m/s (since the fighter's intended path is along the x-axis)
Vresult.y = 515 m/s (since the bullet is traveling along the y-axis)
θ = arctan(515 / 474)
Using a calculator, we find:
θ ≈ 49.7°
The angle at which the bullet actually travels relative to the fighter's intended path is approximately 49.7°.