Math Problem Statement

What is the altitude at the position 10 of the trajectory? Use g = -10 m/s², and round your answers to the nearest hundredth. How far down range is the shell at position 10 of the trajectory?

Solution

To analyze this image and solve for y10y_{10} and x10x_{10}, we need to break down the given projectile motion problem. The following information is provided:

  • Initial velocity v=708m/sv = 708 \, \text{m/s}
  • Launch angle θ=36\theta = 36^\circ
  • Gravitational acceleration g=10m/s2g = -10 \, \text{m/s}^2
  • Position 10 is marked near the end of the trajectory.
  • The time intervals (Δt\Delta t) are given as 4.0 seconds between positions.

We will:

  1. Decompose the velocity into horizontal and vertical components. vx=vcos(θ)v_x = v \cos(\theta) vy=vsin(θ)v_y = v \sin(\theta)

  2. Use the time of flight t10t_{10} and calculate the range x10x_{10}.

  3. Find the altitude y10y_{10} based on the vertical motion equation for position 10.

Let me calculate these values for you. math

Given values

v = 708 # m/s, initial velocity theta = 36 # degrees, launch angle g = -10 # m/s^2, acceleration due to gravity delta_t = 4.0 # seconds between each marked point

Convert angle to radians for trigonometric functions

theta_rad = math.radians(theta)

Horizontal and vertical components of the initial velocity

v_x = v * math.cos(theta_rad) v_y = v * math.sin(theta_rad)

Calculate the total time elapsed until position 10

t_10 = 10 * delta_t

Horizontal range at position 10

x_10 = v_x * t_10

Using vertical motion equation to calculate altitude at position 10

y = v_y * t + 0.5 * g * t^2

y_10 = v_y * t_10 + 0.5 * g * t_10**2

x_10, y_10The calculated values for position 10 of the trajectory are as follows:

  • y10=8646.08my_{10} = 8646.08 \, \text{m} (the altitude)
  • x10=22911.36mx_{10} = 22911.36 \, \text{m} (the range)

Would you like more details or explanations on these calculations?

Here are 5 related questions for further exploration:

  1. How would the trajectory change if the launch angle increased to 45°?
  2. What is the total time of flight for this projectile?
  3. How does changing gravity, such as on another planet, affect the trajectory?
  4. What is the maximum altitude ymaxy_{\text{max}} reached during the trajectory?
  5. How would air resistance modify the range and altitude?

Tip: For projectile motion, the horizontal velocity remains constant (ignoring air resistance), while the vertical velocity is influenced by gravity.

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Math Problem Analysis

Mathematical Concepts

Projectile Motion
Kinematics
Trigonometry

Formulas

v_x = v * cos(θ)
v_y = v * sin(θ)
y = v_y * t + 0.5 * g * t^2
x = v_x * t

Theorems

Kinematic Equations of Motion

Suitable Grade Level

High School (Grades 10-12)