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.)
use the equation θ=(1/2)sin^(-1)〖(gravity x distance)/velocity〗
To solve this problem, we can use the principles of projectile motion and trigonometry. The key is to find the launch angles that would result in the bullet hitting the target at a distance of 91.4 meters.
Let's start by considering the horizontal and vertical components of the bullet's motion.
The horizontal component remains constant throughout the flight because there are no horizontal forces acting on the bullet after it is fired. Therefore, the horizontal distance traveled by the bullet is given by:
d = v₀ * t * cosθ
where d is the horizontal distance, v₀ is the muzzle speed of the bullet, t is the time of flight, and θ is the launch angle.
In this case, the horizontal distance is equal to 91.4 meters. Therefore, we can rewrite the equation as:
91.4 = v₀ * t * cosθ ----(1)
Now, let's consider the vertical component of the bullet's motion. The bullet will follow a parabolic trajectory, and its vertical displacement can be described by the equation:
y = v₀ * t * sinθ - (1/2) * g * t²
where y is the vertical displacement and g is the acceleration due to gravity (assuming a value of 9.8 m/s²).
At the maximum height of the projectile's trajectory, the vertical displacement is zero. This happens when the time is half of the total time of flight (t/2). Therefore, we have:
0 = (v₀ * t/2 * sinθ) - (1/2) * g * (t/2)²
Simplifying this equation gives us:
0 = (sinθ/2) * (v₀ * t - g * t²/2) ----(2)
Combining equation (1) and (2), we can now solve for the launch angle θ.
Since the equation is quadratic in sinθ, we can apply the quadratic formula:
sinθ = [ -b ± √(b² - 4ac) ] / 2a
where a = 1/2, b = -g * t/2, and c = 0.
Plugging in the values, we get:
sinθ = [ -(-g * t/2) ± √((-g * t/2)² - 4 * (1/2) * 0) ] / 2 * (1/2)
Simplifying further, we have:
sinθ = [ g * t/2 ± √(g² * t²/4) ] / 1
sinθ = [ g * t/2 ± (g * t/2) ] / 1
sinθ₁ = g * t/2 + (g * t/2) = g * t
sinθ₂ = g * t/2 - (g * t/2) = 0
Since sinθ cannot be greater than 1, we can discard sinθ₂ = 0. Therefore, the angle θ₂ is never used in target shooting.
We are left with sinθ₁ = g * t. To find the angle θ₁, we need to find the time of flight t.
The time of flight can be found using the equation:
t = 2 * v₀ * sinθ / g
Plugging in the values, we get:
t = 2 * v₀ * (g * t) / g
Simplifying, we have:
1 = 2 * (v₀/g) * t
t = g * v₀ / 2g
t = v₀ / 2
Substituting this back into sinθ₁ = g * t, we get:
sinθ₁ = g * (v₀ / 2)
Using the identity 2 sinθ cosθ = sin 2θ, we can rewrite this as:
2 sin(θ₁) cos(θ₁) = g * (v₀/2)
sin(2θ₁) = g * (v₀/2)
Therefore, we can solve for the angle θ₁ as:
2θ₁ = sin⁻¹(g * (v₀/2))
θ₁ = (1/2) * sin⁻¹(g * (v₀/2))
Finally, we can substitute the given values of v₀ = 392 m/s and g = 9.8 m/s² into the equation to find θ₁:
θ₁ = (1/2) * sin⁻¹(9.8 * (392/2))
θ₁ ≈ 0.466 radians ≈ 26.7 degrees
So, the smaller angle θ₁ is approximately 0.466 radians or 26.7 degrees.
Since we determined earlier that θ₂ is never used in target shooting, we don't need to calculate it.
In conclusion:
(a) Smaller angle: θ₁ ≈ 0.466 radians ≈ 26.7 degrees
(b) Larger angle: Not applicable