Projectile Motion!

Physics 12Mr. Reeves

• A projectiles total velocity is composed of how many components?

• What are these components?

•Velocity Components Review

Vertical (Y) Velocity

Horizontal (X) Velocity

Total Velocity

•Angry Birds!!They’re angry. And following the laws of Physics!

•Today’s Objectives:

Review components of an objects velocity

Introduce the formula for projectile motion

1

2

3 Using the formula, students will be able to solve a projectile motion problem when given initial conditions.

• D = Vi * t + (1/2) *a * t^2

• D = displacement• Vi = Initial velocity• t = Time of travel• a = acceleration

• *Remember, this formula can be applied to both horizontal and vertical components!

PROJECTILE MOTION:

•The motion of an object can be explained using this formula:

Step 1• Record all known values from the problem

Step 2• Use the vertical components to calculate the time the ball

will be in the air

Step 3• Use the time and the horizontal components to calculate

the distance the ball will travel

•Make sure to label units foreach value.

•The better your booking keeping is, the easier your calculations will be!

•Steps to solving a projectile motion problem

• The boy on the tower throws a ball a distance of 20m. At what speed is the ball thrown?

•Lets review projectile motion!

•Solution:First let’s list our known values.

• Horizontal Information Vertical Information• Dx = 20m Dy = 4.9m• vix = ??? viy = 0 m/s• ax = 0 m/s/s ay = 9.8 m/s/s

Dy = Viy*t+(1/2)Ay*t^24.9m = (0 m/s)•t + 0.5•(9.8 m/s/s)•t2

4.9m = (4.9 m/s/s)•t2

1 s2 = t2

t = 1 s

•We can use the vertical components of the ball’s motion to determine the time it will travel.

Dx = Vix*t+(1/2)Ax*t^220m = (Vix)•(1s) + 0.5•(0 m/s/s)•(1s)2

20m = (Vix)•(1s)Vix= 20m/1s = 20m/s

We now use the time which we calculated and our horizontal components to calculate the initial velocity.

• A 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.

•Lets try another!

•Solution:First let’s list our known values.

• Horizontal Information Vertical Information• Dx = ??? Dy = 0.60 m• vix = 2.4 m/s viy = 0 m/s• ax = 0 m/s/s ay = 9.8 m/s/s

<---???--->

|.6m|

2.4 m/s

9.8m/s/s

Dy = Viy*t+(1/2)Ay*t^20.60 m = (0 m/s)•t + 0.5•(9.8 m/s/s)•t2

0.60 m = (4.9 m/s/s)•t2

0.122 s2 = t2

t = 0.350 s

•We can use our vertical components to find time!

Dx = Vix*t+(1/2)Ax*t^2Dx = (2.4 m/s)•(0.3499 s) + 0.5•(0 m/s/s)•(0.3499 s)2

Dx = (2.4 m/s)•(0.3499 s)Dx = 0.84 m

Now we can use our horizontal components to determine where the ball will land.

At the instant a horizontally held rifle is fired, a bullet held at the rifle’s side is released and drops to the ground. Which bullet strikes the ground first?

http://www.youtube.com/watch?v=D9wQVIEdKh8

•Theory

On a separate paper answer these questions:

•Does the time of travel for a projectile depend on the horizontal components or vertical components of its motion? How about the distance it travels?

•What is one thing about today’s lesson that you still have questions about? If there is nothing, what is one thing you learned today?

Have a good day!

•Before you leave!