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Projectile Motion: The most common type of 2-dimensional kinematics problem is the projectile motion problem. A projectile is any massive object travelling through the air without any forces acting upon it other than gravity. In reality air resistance can be a significant factor, however as a first approximation we will assume that air resistance is negligible; that is we will assume that the effects of air resistance are small enough that we can ignore them. What all of this boils down to is: A projectile has a constant acceleration of g=9.80 m s 2 down. We will solve a projectile problem like every other 2-dimensional vector problem. Break the problem into components. List the known quantities in each direction. A neat labelled diagram will help. In any projectile motion problem, we start the problem knowing: a x =0 m s 2 and a y =−9.80 m s 2 ^ y Because a x =0 m s 2 , v x is constant and there is only one simple equation in the x-direction: d x = v x t If you know any 2 of these 3 variables you can solve for the third. The y-direction is a standard 1-dimensional problem with a constant acceleration: a y =−9.80 m s 2 ^ y Just make a list of the 5 variables: a y , d y , v y , v 0 y and t If you know any 3 of these you can find the rest using the 4 key formulas of kinematics. It is important to rmember that time is scalar. This means that time is always the same for both the ^ x and ^ ydirections! In most projectile problems you will use one direction to find time, then use that time to finish solving the other direction.

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Page 1: mcpheesics.weebly.commcpheesics.weebly.com/uploads/6/5/6/9/6569707/n_pr… · Web viewProjectile Motion: The most common type of 2-dimensional kinematics problem is the projectile

Projectile Motion:The most common type of 2-dimensional kinematics problem is the projectile motion problem. A projectile is any massive object travelling through the air without any forces acting upon it other than gravity. In reality air resistance can be a significant factor, however as a first approximation we will assume that air resistance is negligible; that is we will assume that the effects of air resistance are small enough that we can ignore them.

What all of this boils down to is: A projectile has a constant acceleration of g=9.80ms2

down.

We will solve a projectile problem like every other 2-dimensional vector problem. Break the problem into components.

List the known quantities in each direction. A neat labelled diagram will help. In any projectile motion problem, we start the problem knowing:

ax=0ms2

and a y=−9.80 ms2y

Because ax=0ms2

, vx is constant and there is only one simple equation in the x-direction:

d x=v x t

If you know any 2 of these 3 variables you can solve for the third.

The y-direction is a standard 1-dimensional problem with a constant acceleration: a y=−9.80 m

s2y

Just make a list of the 5 variables: a y,d y , v y , v0 y and t

If you know any 3 of these you can find the rest using the 4 key formulas of kinematics.

It is important to rmember that time is scalar. This means that time is always the same for both the x and ydirections!

In most projectile problems you will use one direction to find time, then use that time to finish solving the other direction.

Another helpful fact to remember is that at the vertex of the flight (i.e. the highest point) v y=0m /s.

Page 2: mcpheesics.weebly.commcpheesics.weebly.com/uploads/6/5/6/9/6569707/n_pr… · Web viewProjectile Motion: The most common type of 2-dimensional kinematics problem is the projectile

The setup will look something like:

1. A diagram of the situation:

2. List all of the knowns in the x and y directions.3. Figure out which direction you can solve first.

x : d x= y : a y=−9.80m /s2 d y=12 ( v y+ v0 y) t

vx= d x=v x t v0 y=¿ v y=v0 y+ ay t

t= v y=¿ d y= v0 y t+12a y t

2

d y=¿ v y2 = v0 y2+2 a y d yt=¿

4. Usually you will solve one direction for time and then substitute that time into the other direction. Time is the same for x and y!

Example 1:

A car drives horizontally off of a 13.0m high cliff at 16.2m/s. How far from the base of the cliff will the car crash down in a tragic scene of twisted metal and broken glass?

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Example 2:

A canon ball is fired from the top of a 4.0m high wall. The ball is fired at 128m/s [29o above horizontal]. Find the range of the flight.

\

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Example 3: A tour bus drives off of a cliff. The bus leaves the road at 11m/s [19o below horizontal] and hits the ground 14m from the base of the cliff. Find the height of the cliff.

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