The drawing shows an exaggerated view of a rifle that has been ‘sighted in' for a 91.4-meter target. If the muzzle speed of the bullet is v0 = 392 m/s, there are the two possible angles θ1 and θ2 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. (Hint: The following trigonometric identity may be useful: 2 sinθ cosθ = sin 2θ.) Thanks!
To solve this problem, we can use the horizontal and vertical motion equations to find the two possible angles.
Step 1: Determine the time of flight for the bullet to reach the target.
The horizontal motion equation is given by:
x = v0 * cos(θ) * t
Since the horizontal distance is equal to 91.4 meters, we have:
91.4 = v0 * cos(θ) * t
The vertical motion equation is given by:
y = v0 * sin(θ) * t - (1/2) * g * t^2
We know that the initial vertical displacement is zero at the ground level, so:
0 = v0 * sin(θ) * t - (1/2) * g * t^2
Since the bullet hits the target at the same time it hits the ground, we can set the time t for both equations equal to each other.
v0 * cos(θ) * t = 91.4
v0 * sin(θ) * t - (1/2) * g * t^2 = 0
Step 2: Solve the equations simultaneously.
Solving the first equation for t, we get:
t = 91.4 / (v0 * cos(θ))
Substituting this value into the second equation, we have:
v0 * sin(θ) * (91.4 / (v0 * cos(θ))) - (1/2) * g * (91.4 / (v0 * cos(θ)))^2 = 0
Simplifying the equation, we get:
sin(θ) - (1/2) * g * (91.4 / (v0^2 * cos^2(θ))) = 0
Using the trigonometric identity 2sinθcosθ = sin(2θ), we can rewrite the equation as:
2sin(θ)cos(θ) - (1/2) * g * (91.4 / (v0^2 * cos^2(θ))) = 0
Step 3: Solve for the angles.
Rearranging the equation, we have:
2sin(θ)cos(θ) = (1/2) * g * (91.4 / (v0^2 * cos^2(θ)))
Multiplying both sides of the equation by 2 and rearranging again, we get:
sin(2θ) = (g * 91.4) / (v0^2 * cos^2(θ))
Using the trigonometric identity sin(2θ) = 2sin(θ)cos(θ), we can rewrite the equation as:
2sin(θ)cos(θ) = (g * 91.4) / (v0^2 * cos^2(θ))
Cancelling out the common terms, we have:
cos(θ) = (g * 91.4) / (v0^2 * cos^2(θ))
Since cos^2(θ) is never negative, we can divide both sides by cos^2(θ):
1 = (g * 91.4) / (v0^2)
Simplifying the equation, we get:
v0^2 = g * 91.4
Now we can solve for v0:
v0 = √(g * 91.4)
Substituting the given value of v0 = 392 m/s, we have:
392 = √(g * 91.4)
Squaring both sides of the equation, we get:
153,664 = g * 91.4
Solving for g, we have:
g = 153,664 / 91.4
Step 4: Calculate the angles.
Now we can use the value of g to find the angles θ1 and θ2.
θ1 = sin^(-1)((91.4 * g) / (v0^2))
θ2 = 90° - θ1
Substituting the given values, we can calculate the angles using a calculator.
θ1 ≈ 0.048 radians (smaller angle)
θ2 ≈ 89.952 radians (larger angle)
Converting radians to degrees, we get:
θ1 ≈ 2.757° (smaller angle)
θ2 ≈ 5141.588° (larger angle)
The smaller angle is θ1 ≈ 2.757°, and the larger angle is θ2 ≈ 5141.588°. Since the larger angle is extremely large, it is not used in target shooting.
To find the values of the two angles θ1 and θ2, we can use the equations of projectile motion. Let's assume the rifle is fired from a height of h = 0 meters.
First, we can break down the initial velocity v0 into its horizontal and vertical components. The horizontal component, Vx, remains constant throughout the motion and is given by Vx = v0 * cosθ. The vertical component, Vy, changes due to the acceleration due to gravity and is given by Vy = v0 * sinθ - g * t, where g is the acceleration due to gravity (approximately -9.8 m/s^2) and t is the time of flight.
The horizontal range, R, is given by R = Vx * t. In this case, the range is 91.4 meters.
Now, we need to find the time of flight. Since the maximum height is not given, we assume that the bullet hits the target at the same height it was fired from. Thus, the equation becomes h = Vy * t - (1/2) * g * t^2, where h = 0 and we solve for t.
0 = v0 * sinθ * t - (1/2) * g * t^2
Now, we can substitute for t using the equation for R:
t = R / Vx
Substituting this into the previous equation, we get:
0 = (v0 * sinθ / Vx) * (R / Vx) - (1/2) * g * (R / Vx)^2
Simplifying this equation, we get:
0 = 2 * sinθ * cosθ - (g * R^2) / (v0^2 * cos^2θ)
Using the given trigonometric identity, 2 sinθ cosθ = sin 2θ, we can rewrite the equation as:
0 = sin 2θ - (g * R^2) / (v0^2 * cos^2θ)
Now, we solve this equation numerically to find the values of θ1 and θ2. We can use a scientific calculator, computer program, or spreadsheet software to find the solutions.
Once we find the values of θ1 and θ2, we compare them to determine which one is larger. The larger angle is the one that is never used in target shooting.
Note: Since the values of g and v0 are not provided, we cannot calculate the actual angle values. However, by following the steps outlined above and substituting the given values, you should be able to find the answer.