Projectile MotionThe motion of an object that is thrown and moves along a curved path

through the influence of gravity, only.

1. The path is the resultant of two other motions.A. Consider horizontal velocity first. Roll a ball along the floor.•Where does its speed come from? The initial velocity you

gave it.• If the floor is smooth and level, it keeps right on rolling at

that same speed for a long time.•Without friction, vf = vi

B. Now consider vertical velocityThrow the ball straight up in the air.Where did the original motion come from? You.But does the velocity remain constant?

This time, there is an unbalanced force, gravity, also acting on the ball. acceleration due to gravity (g) changes the velocity over time.The ball, no matter how strong you are, will eventually stop.

vf will = 0.

• C. Now roll the ball along a table and let it go off the end of the table.• Note the path of the ball as it ‘flies’ off the end.

• It is an arc, or half a projectile path:

Why doesn’t it fall straight down? Or fly straight out?• It has motion in two directions:• It has the initial horizontal motion you gave it.

It has vertical motion because gravity was free to bring it down.

The initial vertical velocity was zero. But what was the final vertical velocity?The final horizontal velocity?

• Each graph represents a ball flying off the table.• On each

graph, number the x and y axes beginning at zero.• Use a 1 m

interval on both axes.

• 1. Number each of the graphs. Begin each axis at zero, and use 1 m interval.• 2. Draw a point

on the line for the horizontal 3 m mark.• 3. Draw a tangent

to that point.• 4. Draw the

horizontal and vertical component.

• 5. Are the 4 arcs identical? Or are some wider than high? Flatter? Steeper? Which of the graphs has the greatest HORIZONTAL distance? __________ • Which has the greatest VERTICAL distance? ___________

• 6. For each of the arcs, we will find the vertical and horizontal velocities.

a. First, we will ONLY look at the vertical axis. what is the ‘d’ value vertically for the arc? ______ Do we know acceleration? (hint; yes) Do we know vi? (yes)

Solve for time:Equation to use: d = (vi * t) + ½ a t2 Plug in the numbers: Ex. 6m = 0 + 5m/

1.2 = 1.1 s = t

• b. To find the horizontal velocity, use the time from the previous problem and ‘d’ from the horizontal axis.

• Then, use

6m = v * 1.1 s ; Vh = 5.5 m/s

Repeat these steps for the other 3 graphs.

What else can we calculate from the graphs?

•What was the vertical velocity at impact?vf = vi + a*t : vf = 10m /* t

We could also go to any point on the arc and find the vertical velocity at that point.We could find the actual velocity of the ball as the resultant of the horizontal and vertical velocity.

• D. Vectors will be used to demonstrate the motion.• The actual path of the ball can be shown as a line tangent from one

point on the curve:The blue line is the resultant of the horizontal and vertical velocities.

• The ball was never given a ‘push’ at that angle. So how did it happen?• The motion of the ball is the resultant of two component vectors at

that point.•

The red line is the horizontal component.The magenta line is the vertical component.

• The horizontal and vertical vectors are independent of each other.• But the ball can only move in one direction at any moment in time.

• So, the actual motion of the ball is a combination of the two component motions.• The path of the ball is the resultant of the two motions.

• Notice the vector diagram is similar to the parallelogram vectors we’ve looked at before.

The black lines are the horizontal and vertical components.The blue and green ‘finish’ the parallelogram.The red line is the resultant.

At each point on the path the cannon ball travels, it has two motions.

Initially, its vertical velocity = 0 m/s. Its horizontal velocity is the speed at which it left the cannon.

• But at each second that it moves, the vertical velocity increases because of the acceleration due to gravity. What happens to the horizontal velocity?

• Two balls dropped at the sametime. The yellow ball was thrownslightly to the right as it was dropped.The red was just dropped.What do they have in common?

Using trig to find the values

How can we resolve component and resultant vectors?

: In this class, we have used:

1. Pythagorean theorem: = (in which ‘c’ is the hypotenuse)2. Draw scale vectors and use the protractors (very time-

consuming)3. Use trig: SOH, CAH, TOA.

A review of Trig functions:

Which is the opposite? Which is the adjacent? It depends on the angle you are considering.

For this triangle, I would be doing all the work using angle ‘x’.So; SOH = sin ‘x’ = B/C CAH = cos ‘x’ = A/C TOA = tan ‘x’ = B/A

x

Z

For the angle ‘z’;

SOH = sin ‘Z’ = A/C CAH = cos ‘Z’ = B/C TOA = tan ‘Z’ = A/B

Now, for a fourth way:

• The only triangles we’ll use (text p. 30-31)

• The right triangle: = since a and b are the same: = 2 ; c = x a ‘H’ = 1.414 x length of either side.

• The other triangle to consider: the famous ‘3, 4, 5’ triangle: learn this triangle!!!!

.We can solve the problems with trig;

Or, do simple cross-multiplying of ratios.

1. Learn the relationships:

The side opposite = 3 /5 Hypotenuse.

The side opposite = 4/5 Hypotenuse.

2. Do the math: example: If the hypotenuse is 20 m/s;

then the value of the other sides of thetriangle are: 4/5 x 20m/s; and 3/5 x 20m/s.

• Name _____________________ Draw the triangle and label what you know. Then solve for the horizontal and vertical velocities using either trig functions or what you know about these two types of triangles.1. A ball has a velocity of 10m/s at an angle with the ground of 37

degrees.2. A duck lands at an angle of 53 degrees to the water with a speed of

20 m/s.3. A batter hits a homerun at an angle of 45 degrees and a speed of

40m/s. 4. What would be the horizontal and vertical speeds of the ball from

question #3 at the very top of the arc? Explain.