
Objective:
By the end of this module you should be able to:
 Solve problems involving vertical free fall using equations for uniformly accelerated motion.
 Recognize that projectile motion can be analyzed by considering the horizontal and vertical components of the motion separately.
 Apply kinematic equations for constant velocity to analyze the horizontal motion of a projectile.
 Apply kinematic equations for uniform acceleration to analyze the vertical motion of a projectile.

Introduction:
Projectiles can also be launched at an angle to the Earth's surface. When this occurs, the velocity in the vertical direction (viy) is no longer equal to zero. There is velocity in both the horizontal and vertical directions. And the shape of the trajectory will now include both sides of the parabola: as the projectile moves up and as it moves down.
However, the analysis of the motion is virtually the same as before. The only thing that is different is we now need to determine the initial vertical and horizontal components of the initial velocity.
To determine the horizontal and vertical components of the initial velocity we will use trigonometric identities: 


All other equations remain the same: 
In the horizontal direction:

In the vertical direction:

The diagram below shows the entire path of a projectile launched at an angle. We can see the velocity at certain points along the trajectory; as well as the corresponding horizontal and vertical components of the velocity.
Notice that the horizontal component is of equal magnitude at any point along the trajectory. Again, because we are ignoring air resistance, there is no external force in the horizontal direction. So the horizontal component of velocity will remain equal at any point along the projectile's path.
As always, due to gravity, the vertical component of velocity changes along its trajectory. Here are some of the special points to made note of:
 Initially the vertical component of velocity is large and acting in the upwards direction.
 As the projectile continues to move upwards, the vertical velocity decreases at a rate of 9.81 meters per second squared (acceleration due to gravity).
 When the projectile reaches the top of the trajectory (its maximum height), the vertical component of velocity is zero. At this point, the velocity is equal to the horizontal component of velocity.
 As the projectile moves downward, the vertical component of velocity increases in the downward direction; again, at a rate of 9.81 meters per second squared.
Another feature to take note of is that the speed is equal at equal heights. The only difference is the direction.

Example:
A bouncing ball leaves the ground with a velocity of 4.36 m/s at an angle of 81 degrees above the horizontal.
a) How long did it take the ball to land?
b) How high did the ball bounce?
c) What was the ball's range?

Activities:
Reading:
Read the following links:
Take notes as you read through the sections.
Simulation:
Using the worksheet below, go through the following simulation activity. When you get to the simulation web site, use the menu at the left of the screen to choose: Projectile Motion.

Practice:
Answer the following theory questions. Email your answers to your "homework buddy" and send them feedback on their responses. Remember to cc a copy to me. Once I receive your answers I will email you back with the solutions.
 A projectile is fired straight upward at 100 m/s. How fast is it moving at the instant it reaches the top of its trajectory?
 What would the answer to question 1 be if the projectile is fired upward at 45 degrees instead?
 At what angle should a slingshot be oriented for maximum horizontal range? For maximum altitude?
 Neglecting air resistance, if you throw a ball straight up with a speed of +20 m/s, what will be its speed when you catch it?
 Neglecting air resistance, if you throw a baseball at 20 m/s to your friend who is on first base, will the catching speed be greater than, equal to, or less than 20 m/s? What about if air resistance is a factor?
Answer the following question. The solution is given below, but make a fair attempt to do the question on your own. If you are having difficulty or do not understand the solution you can either email your "homework buddy" or myself.
 A golf ball is hit with a velocity of 24.5 m/s at 35.0 degrees above the horizontal. Find a) the range of the ball, and b) the maximum height of the ball.
 A player kicks a football from ground level at 27.0 m/s at an angle of 30.0 degrees above the horizontal. Find a) its “hang time” (time that the ball is in the air), b) the distance the ball travels before it hits the ground, and c) its maximum height.
 After retrieving her ball from the roof, a girl throws the ball off of the roof with a speed of 5.6 m/s at an angle of 35 degrees above the horizontal. The ball is thrown from a height of 11 m above the ground. Find a) the time the ball is in the air, b) the range of the ball, c) the maximum height of the ball, and d) the speed of the ball as it strikes the ground.

