Problem Type 2:

A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally. Upon reaching the peak, the projectile falls with a motion that is symmetrical to its path upwards to the peak. Predictable unknowns include the time of flight, the horizontal range, and the height of the projectile when it is at its peak. Examples of this type of problem are

a. A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the football. b. A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.

The second problem type will be the subject of the next part of Lesson 2. In this part of Lesson 2, we will focus on the first type of problem - sometimes referred to as horizontally launched projectile problems. Three common kinematic equations that will be used for both type of problems include the following:

Equations for the Horizontal Motion of a Projectile

The above equations work well for motion in one-dimension, but a projectile is usually moving in two dimensions - both horizontally and vertically. Since these two components of motion are independent of each other, two distinctly separate sets of equations are needed - one for the projectile's horizontal motion and one for its vertical motion. Thus, the three equations above are transformed into two sets of three equations. For the horizontal components of motion, the equations are

Of these three equations, the top equation is the most commonly used. An application of projectile concepts to each of these equations would also lead one to conclude that any term with ax in it would cancel out of the equation since ax= 0 m/s/s.

Equations for the Vertical Motion of a Projectile

For the vertical components of motion, the three equations are

In each of the above equations, the vertical acceleration of a projectile is known to be -9.8 m/s/s (the acceleration of gravity). Furthermore, for the special case of the first type of problem (horizontally launched projectile problems), viy = 0 m/s. Thus, any term with viy in it will cancel out of the equation. The two sets of three equations above are the kinematic equations that will be used to solve projectile motion problems.

Solving Projectile Problems

To illustrate the usefulness of the above equations in making predictions about the motion of a projectile, consider the solution to the following problem.

ExampleA pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.| The solution of this problem begins by equating the known or given values with the symbols of the kinematic equations - x, y, vix, viy, ax, ay, and t. Because horizontal and vertical information is used separately, it is a wise idea to organized the given information in two columns - one column for horizontal information and one column for vertical information. In this case, the following information is either given or implied in the problem statement: Horizontal Information| Vertical Information|

x = ???vix = 2.4 m/sax = 0 m/s/s| y = -0.60 mviy = 0 m/say = -9.8 m/s/s| As indicated in the table, the...