Chapter 3: Problem 226
The following problems consider launching a cannonball out of a cannon. The cannonball is shot out of the cannon with an angle \(\theta\) and initial velocity \(\mathbf{v}_{0} .\) The only force acting on the cannonball is gravity, so we begin with a constant acceleration \(\mathbf{a}(t)=-g \mathbf{j}\) Find the velocity vector function \(\mathbf{v}(t)\)
Short Answer
Step by step solution
Initial Setup
Separate Components
Integrating the Acceleration: Vertical Component
Integrating the Acceleration: Horizontal Component
Constructing the Velocity Vector
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Newton's Second Law
Velocity Vector
- Horizontal Component \( v_x(t) \): Influenced solely by the initial velocity and angle of projection. Since no horizontal forces act (assuming no air resistance), it remains constant. \( v_x(t) = v_0 \cos(\theta) \).
- Vertical Component \( v_y(t) \): Changes over time due to gravitational acceleration. It starts from an initial value and decreases linearly with time because of gravity: \( v_y(t) = -gt + v_0 \sin(\theta) \).
Integration
- Vertical Integration: The acceleration due to gravity is \( -g \). The differential equation \( \frac{dv_y}{dt} = -g \) integrates to \( v_y(t) = -gt + C \). The integration constant \( C \) is determined using initial conditions: \( v_{0y} = v_0 \sin(\theta) \).
- Horizontal Integration: Here, \( \frac{dv_x}{dt} = 0 \) implies no horizontal acceleration. Integration gives a constant, \( v_x(t) = C \), which is \( v_0 \cos(\theta) \) by initial conditions.
Gravity
- It causes the vertical velocity component \( v_y(t) \) to change over time, while horizontal motion remains unaffected.
- It ensures that after reaching a certain height, the projectile begins descending back to the ground.
- Gravity dictates the parabolic shape of the projectile's trajectory.