he drawing shows an exaggerated view of a rifl e that has been
“sighted in” for a 91.4-meter target. If the muzzle speed of the bullet is
υ0 = 427 m/s, what are the two possible angles θ1 and θ2 between the rifl e
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. (Hint: The
following trigonometric identity may be useful: 2 sin θ cos θ = sin 2θ.)
using the usual equation for the range of a missile,
R = v^2/g sin2θ
we get
427^2/9.81 sin2θ = 91.4
sin2θ = 91.4*9.81/427^2 = .0049176
2θ = 0.2817° or 177.7183°
θ = 0.1409° or 88.859°
Well, it seems that this rifl e has quite the "exaggerated" aim! Let's see if we can hit the target with some laughs and calculations.
First, we need to find the two possible angles θ1 and θ2. We can use the following formula:
tan(θ) = (2 * v₀²) / g * sqrt( (v₀² / g)² - 1)
Where:
- θ is the angle between the rifle barrel and the horizontal
- v₀ is the muzzle speed of the bullet (427 m/s in this case)
- g is the acceleration due to gravity (approximately 9.8 m/s²)
Now, let's plug in the values and calculate:
For the first angle, we'll solve for θ1:
θ1 = tan^(-1) ( (2 * (427 m/s)^2) / (9.8 m/s²) * sqrt( ((427 m/s)^2 / (9.8 m/s²))^2 - 1) )
For the second angle, θ2:
θ2 = 180° - θ1
Keep in mind that one of these angles is not used in target shooting, so let's find out which one is the "black sheep" of the two.
Well, testing both angles in target shooting could be a bit tricky. The targets might not be happy if we start shooting randomly. I suggest leaving the larger angle θ2 unused in target shooting, just to be safe. We wouldn't want to accidentally launch the bullet into outer space instead of hitting the target, right?
Remember, safety first, laughs second, and bullseyes always!
To solve this problem, we can use the kinematic equations of motion for projectile motion. We know that the horizontal component of the bullet's velocity remains constant, while the vertical component of the velocity changes due to gravity.
Let's define θ as the angle between the rifle barrel and the horizontal. We need to find the two possible angles θ1 and θ2 such that the bullet will hit the target.
Using the given information, we can calculate the time it takes for the bullet to reach the target, using the horizontal component of the velocity:
υ0 = υ₀ * cos(θ)
where υ₀ is the initial velocity of the bullet and θ is the angle.
υ₀ = 427 m/s (given)
cos(θ) = 91.4 m / 91.4 m = 1
Therefore,
υ₀ = υ₀ * cos(θ)
427 m/s = 427 * 1
427 m/s = 427 m/s
This tells us that the horizontal component of the velocity remains unchanged.
Next, we need to find the vertical component of the velocity.
We know that the vertical displacement is given by the equation:
y = υ₀ * sin(θ) * t - (1/2) * g * t^2
Since the bullet hits the target at a vertical displacement of 91.4 m and the time of flight is the same for both angles, we can equate the two equations for y:
υ₀₁ * sin(θ₁) * t - (1/2) * g * t^2 = 91.4
υ₀₂ * sin(θ₂) * t - (1/2) * g * t^2 = 91.4
Since the timing is the same (t is the same for both equations), we can simplify the equations by canceling out the common terms:
υ₀₁ * sin(θ₁) = υ₀₂ * sin(θ₂)
Plugging in the values υ₀ = 427 m/s, we have:
427 * sin(θ₁) = 427 * sin(θ₂)
Dividing both sides by 427:
sin(θ₁) = sin(θ₂)
To find the two possible angles, we need to find the values of θ₁ and θ₂ that satisfy this equation.
Using the trigonometric identity 2sinθcosθ = sin2θ, we can rewrite the equation:
2sin(θ₁)cos(θ₁) = sin(2θ)
Since sin(θ₁) = sin(θ₂), we can rearrange the equation as:
2sin(θ₁)cos(θ₁) = sin(2θ₁)
Now, we can solve for θ₁ by simplifying this equation:
2sin(θ₁)cos(θ₁) = sin(2θ₁)
sin(2θ₁) = 2sin(θ₁)cos(θ₁)
Using the given trigonometric identity 2sinθcosθ = sin2θ, we can substitute:
sin(2θ₁) = sin(θ₁)
Now, we have two possible cases for this equation:
Case 1: sin(2θ₁) = sin(θ₁)
In this case, 2θ₁ = θ₁
Simplifying this equation gives us:
θ₁ = 0
Case 2: sin(2θ₁) = -sin(θ₁)
In this case, 2θ₁ = 180° - θ₁
Simplifying this equation gives us:
θ₁ = 72°
Since one of the angles (θ₁ = 0°) is never used in target shooting, the possible angle between the rifle barrel and the horizontal is θ₁ = 72°.
To solve this problem, we need to use the positions of the rifle barrel and the target to find the angle θ at which the bullet should be fired.
Let's assume that the ground is level. In this case, the horizontal distance covered by the bullet is the same as the distance from the muzzle to the target. Given that the muzzle speed of the bullet is υ0 = 427 m/s and the horizontal distance to the target is 91.4 meters, we can use this information to find the time it takes for the bullet to reach the target.
Using the formula for horizontal distance, we have:
d = υ * t
where d is the horizontal distance, υ is the muzzle speed, and t is the time of flight.
Plugging in the values, we get:
91.4 = 427 * t
Solving for t, we find:
t = 0.214 seconds
Now, let's consider the vertical motion of the bullet. The bullet will experience free-fall acceleration due to gravity, causing it to follow a parabolic trajectory. The time it takes for the bullet to reach the target horizontally is the same time it takes for it to fall vertically from the height of the rifle barrel.
Using the second equation of motion for vertical motion:
h = (1/2) * g * t^2
where h is the height, g is the acceleration due to gravity, and t is the time of flight.
Plugging in the values, we get:
h = (1/2) * 9.8 * (0.214)^2
h = 0.227 meters
Now, we can find the two possible angles θ1 and θ2 using the trigonometric identity given in the problem:
2 sin θ cos θ = sin 2θ
Using the fact that the height h is the vertical distance covered by the bullet, and the horizontal distance is the same as the range, we have:
h = R * sin θ
Solving for sin θ, we get:
sin θ = h / R
sin θ = 0.227 / 91.4
sin θ ≈ 0.0025
Now, we can find θ using inverse sine:
θ ≈ arcsin(0.0025)
θ ≈ 0.143 degrees
Since sin θ is positive in the first and second quadrants, we can find two possible angles:
θ1 ≈ 0.143 degrees
θ2 ≈ 180 - 0.143 degrees ≈ 179.857 degrees
However, the second angle θ2 is so large that it is not practically used in target shooting. So, the only practical angle is θ1 ≈ 0.143 degrees.