Math Problem Statement

Gambar di atas menunjukkan pandangan dari sebuah senapan yang telah 'diarahkan' untuk mengenai target pada jarak 91.4 meter. Jika kecepatan peluru adalah v0 = 427 m/s, berapakah dua sudut yang mungkin, yaitu sudut θ1 dan θ2, antara laras senapan dan garis horizontal sehingga peluru dapat mengenai target? Berapakah waktu yang dibutuhkan peluru untuk mencapai target, dengan menggunakan 2 sudut tersebut?

Solution

The problem provided involves projectile motion, specifically asking to determine two possible angles of projection θ1\theta_1 and θ2\theta_2 such that the projectile hits a target 91.4 meters away, given the initial velocity v0=427m/sv_0 = 427 \, \text{m/s}. It also asks to compute the time it takes for the projectile to reach the target for each angle.

Let's break down the two parts:

Part a: Determining the angles of projection (θ1\theta_1 and θ2\theta_2)

For projectile motion, the horizontal range RR is given by the formula: R=v02sin(2θ)gR = \frac{v_0^2 \sin(2\theta)}{g} where:

  • R=91.4mR = 91.4 \, \text{m} is the range,
  • v0=427m/sv_0 = 427 \, \text{m/s} is the initial velocity,
  • g=9.8m/s2g = 9.8 \, \text{m/s}^2 is the acceleration due to gravity,
  • θ\theta is the angle of projection.

Rearranging this equation to solve for θ\theta: sin(2θ)=Rgv02\sin(2\theta) = \frac{R \cdot g}{v_0^2}

Substitute the known values: sin(2θ)=91.4×9.84272\sin(2\theta) = \frac{91.4 \times 9.8}{427^2}

We can solve for 2θ2\theta and then find θ1\theta_1 and θ2\theta_2, as sin(2θ)\sin(2\theta) will give two possible angles.

Part b: Time of flight for each angle

The time of flight TT for a projectile is given by: T=2v0sin(θ)gT = \frac{2v_0 \sin(\theta)}{g} We'll calculate the time for both θ1\theta_1 and θ2\theta_2 obtained in part (a).

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

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

For the time of flight:

  • At θ10.14\theta_1 \approx 0.14^\circ, the time of flight is approximately 0.214 seconds.
  • At θ289.86\theta_2 \approx 89.86^\circ, the time of flight is approximately 87.14 seconds.

Would you like more details or have any questions? Here are some follow-up questions to deepen understanding:

  1. Why are there two possible angles for the same target range?
  2. How does the angle affect the time of flight in projectile motion?
  3. Can you explain how trigonometric identities were used in this solution?
  4. What would happen to the angles if the target were further away?
  5. How does changing the initial velocity affect the angles and time?

Tip: Always remember that in projectile motion, a higher angle gives a higher trajectory, but it also increases the time of flight significantly!

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Projectile Motion
Physics

Formulas

R = (v0^2 * sin(2θ)) / g
T = (2v0 * sin(θ)) / g
sin(2θ) = 2 * sin(θ) * cos(θ)

Theorems

Projectile Range Equation

Suitable Grade Level

Grades 11-12 (High School Physics)