Math Problem Statement

The figure shows a rifle that has been 'sighted in' for a 91.4-meter target. If the muzzle speed of the bullet is v₀ = 427 m/s, what 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.

Solution

This problem involves projectile motion where we need to calculate the two possible angles, θ1\theta_1 and θ2\theta_2, at which a bullet with an initial velocity v0=427m/sv_0 = 427 \, \text{m/s} can be shot to hit a target 91.4 meters away.

The general equation for the horizontal range RR in projectile motion is given by:

R=v02sin(2θ)gR = \frac{v_0^2 \sin(2\theta)}{g}

Where:

  • R=91.4mR = 91.4 \, \text{m} (the horizontal range to the target),
  • v0=427m/sv_0 = 427 \, \text{m/s} (initial velocity),
  • g=9.8m/s2g = 9.8 \, \text{m/s}^2 (acceleration due to gravity),
  • θ\theta is the angle at which the bullet is shot.

Step 1: Solve for sin(2θ)\sin(2\theta)

We can rearrange the equation to solve for sin(2θ)\sin(2\theta):

sin(2θ)=gRv02\sin(2\theta) = \frac{gR}{v_0^2}

Substituting the values:

sin(2θ)=9.8×91.44272\sin(2\theta) = \frac{9.8 \times 91.4}{427^2}

I'll compute this value first.We have sin(2θ)=0.00491\sin(2\theta) = 0.00491.

Step 2: Find the two possible values for 2θ2\theta

Since sin(2θ)\sin(2\theta) gives two possible angles for 2θ2\theta (due to the periodic nature of the sine function), we use:

2θ1=sin1(0.00491)2\theta_1 = \sin^{-1}(0.00491) 2θ2=1802θ12\theta_2 = 180^\circ - 2\theta_1

I'll calculate these values now.The two possible values for 2θ2\theta are:

  • 2θ10.282\theta_1 \approx 0.28^\circ
  • 2θ2179.722\theta_2 \approx 179.72^\circ

Step 3: Solve for θ1\theta_1 and θ2\theta_2

Now, divide these by 2 to get the two possible angles θ1\theta_1 and θ2\theta_2:

θ1=2θ12,θ2=2θ22\theta_1 = \frac{2\theta_1}{2}, \quad \theta_2 = \frac{2\theta_2}{2}

I'll calculate these angles.The two possible angles are:

  • θ10.14\theta_1 \approx 0.14^\circ
  • θ289.86\theta_2 \approx 89.86^\circ

One of these angles (89.8689.86^\circ) is so steep that it is impractical for target shooting, which is why it is typically not used.

Would you like more details or further clarification?

5 Related Questions:

  1. How does air resistance affect the actual trajectory in real-world projectile motion?
  2. What are the effects of varying the initial velocity on the range of the projectile?
  3. Can you explain how to calculate the maximum height of the projectile?
  4. How would you determine the time of flight for the projectile at each angle?
  5. How would this calculation change if the target were at a different height?

Tip:

In projectile motion, always check whether you need to consider both the horizontal and vertical components of motion separately.

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Math Problem Analysis

Mathematical Concepts

Projectile Motion
Trigonometry
Kinematics

Formulas

R = (v₀² * sin(2θ)) / g
sin(2θ) = gR / v₀²
2θ₁ = sin⁻¹(0.00491)
2θ₂ = 180° - 2θ₁

Theorems

Range formula in projectile motion

Suitable Grade Level

Undergraduate Physics or Advanced High School Physics