In projectile motion, the horizontal and the vertical components of the motion are treated separately. A projectile moves both horizontally and vertically. Its horizontal motion is constant. Its vertical motion is affected by the acceleration due to gravity. The only variable shared by both types of motion is time. Every point on the trajectory is the vector sum of the horizontal and the vertical components of the velocity.
An object projected horizontally (projected perfectly parallel to the surface) will reach the ground in the same time as an object dropped vertically. Since speed at any point in a trajectory is the vector sum of the horizontal and vertical velocity components at that point, the projected object will have a greater speed when it strikes.
The maximum range for a given initial velocity is obtained when the angle of projection is 45°.
Equations that are used to describe the horizontal and vertical motion:
where d is the vertical distance, vi is the initial vertical velocity, and vf is the final vertical velocity
We will be working two types of problems.
In this case, q = 0°. Thus, cos q = 1 and sin q = 0 which makes the initial vertical velocity component equal to zero and the horizontal velocity component equal the the horizontally projected velocity.
At pointB, its vertical velocity component is zero. The acceleration at all points is -9.8 m/s2. At point B, the object is accelerating (even though its vertical velocity is zero), because its direction is changing. At all points in the trajectory, the horizontal velocity component remains the same (in the absence of air friction). We will work these problems by finding total time in the air using the second acceleration formula. vf at point C is numerically equal to vi at point A but of the opposite sign. Remember, velocity is a vector quantity! Once we have total time, we can use d = v t to find the horizontal range, knowing that this uses the horizontal velocity componet. We can use the last acceleration formula to predict the maximum vertical displacement (working from points B and D).
A formula can be derived for the horizontal range:
A virtual lab in which the user can control the angle of the cannon. Cursor interrogation allows maximum vertical height and total horizontal range to be determined. A target is provided for amusement.
These two types of problems can also be solved graphically using parametric equations on the graphing calculator. If you are interested in learning how to do this, it is explained in the link below.
Graphing calculator solution to problems
One of the classic questions in physics is this: What should a monkey in a tree do when a gun pointed at him fires -- jump down or stay where it is? A graphing calculator solution to this problem is found at Monkey and Hunter Problem
I like to use the Greek letter, n, as the symbol for frequency. This can be confusing to students since its appearance is similar to the letter v. It can also be represented using the symbol f
characteristics of a simple pendulum:
Where l is the length of the pendulum and g is the acceleration due to gravity at that point.
For a pendulum, speed is zero and acceleration is a maximum at the point of maximum displacement (point A). For a pendulum, speed is a maximum and acceleration is zero at the equilibrium point (point B).
acceleration involves a change in speed and/or direction; it is caused by an unbalanced force
in circular motion, the object moves at constant speed but is accelerating because its direction is constantly changing
Uniform Circular Motion
You can use your graphing calculator to determine how the magnitude of the centripetal force varies the speed with which the object is swung in the horizontal circle, the mass of the object, or the radius of the horizontal circle.
Kepler's Laws of Planetary Motion
Motion in Two Dimensions Homework
Periodic Motion Homework
Motion in Two Dimensions Sample Problems