a rifle that has been ‘sighted in' for a 91.4-meter target. If the muzzle speed of the bullet is v0 = 328 m/s, there are the two possible angles è1 and è2 between the rifle barrel and the horizontal such that the bullet will hit the target. One of these angles is so large that it is never used in target shooting. Give your answers as A) the smaller angle and (b) the larger angle.
To find the two possible angles, we can use the following equation:
d = (v0^2 * sin(2θ)) / g
where:
- d is the target distance (91.4 meters)
- v0 is the muzzle speed of the bullet (328 m/s)
- θ is the angle between the rifle barrel and the horizontal
- g is the acceleration due to gravity (9.8 m/s^2)
Let's solve for θ.
1. For the smaller angle (θ1):
Rearranging the formula, we have:
sin(2θ1) = (d * g) / v0^2
Plugging in the values:
sin(2θ1) = (91.4 * 9.8) / (328^2)
sin(2θ1) = 0.0277
Taking the inverse sine:
2θ1 = arcsin(0.0277)
2θ1 ≈ 0.468 radians
Dividing by 2 to find θ1:
θ1 ≈ 0.234 radians
2. For the larger angle (θ2):
sin(2θ2) = -1 * (d * g) / v0^2
Plugging in the values:
sin(2θ2) = -1 * (91.4 * 9.8) / (328^2)
sin(2θ2) = -0.0277
Taking the inverse sine:
2θ2 = arcsin(-0.0277)
2θ2 ≈ -0.468 radians
Dividing by 2 to find θ2:
θ2 ≈ -0.234 radians
Since the angle cannot be negative, we discard θ2 as it is the larger but unused angle in target shooting.
Therefore, the answers are:
A) The smaller angle (θ1) ≈ 0.234 radians
B) The larger angle (θ2, not used in target shooting) is discarded.
To find the two possible angles, we can use the range formula for projectile motion:
R = (v0^2 * sin(2θ)) / g
Where:
R is the range, which is 91.4 meters in this case.
v0 is the muzzle speed of the bullet, which is 328 m/s.
θ is the angle between the rifle barrel and the horizontal.
g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
Let's solve for the two possible angles:
For the smaller angle (θ1):
91.4 = (328^2 * sin(2θ1)) / 9.8
sin(2θ1) = (91.4 * 9.8) / (328^2)
sin(2θ1) = 0.020854
To find θ1, we take the inverse sine of both sides:
2θ1 = sin^(-1)(0.020854)
θ1 = 0.5 * sin^(-1)(0.020854)
θ1 ≈ 0.193 degrees
For the larger angle (θ2):
91.4 = (328^2 * sin(2θ2)) / 9.8
sin(2θ2) = (91.4 * 9.8) / (328^2)
sin(2θ2) = 0.020854
Again, taking the inverse sine of both sides:
2θ2 = sin^(-1)(0.020854)
θ2 = 0.5 * sin^(-1)(0.020854)
θ2 ≈ 0.193 degrees
The smaller angle (θ1) and the larger angle (θ2) between the rifle barrel and the horizontal are approximately the same, both around 0.193 degrees. However, it is mentioned that the larger angle is never used in target shooting. So, the answer would be:
(a) The smaller angle: 0.193 degrees
(b) The larger angle: Not used in target shooting