Projectile Motion

Examples of Projectile Motion:

•A cannon ball•Throwing a baseball•A ball rolling off a table•Throwing a rock

The list can go on and on.

What is a projectile?

•A projectile is an object upon which the only force acting on it is gravity.

•A projectile is any object that once projected continues in motion by its own inertia and is influenced ONLY by gravity.

Common misconceptions

•Many students ask “How can an object be moving upward if the only force acting on it is gravity?”

•Trust Newton•A force is NOT necessary to keep an object in motion•A force is only necessary to keep an object in

ACCELERATION

•A projectile has two vector components: horizontal and vertical (x and y)

•These two components of motion must be discussed separately

•Acceleration due to gravity (g) only affects the vertical motion of the projectile.

•The horizontal motion is unaffected by gravity.

To Summarize

Non- Horizontally Launched

• If a projectile is launched non-horizontally:

•With gravity, the object falls with a parabolic trajectory or path

•The projectile still moves the same horizontal distance in each second travelled as it did when gravity was NOT a factor

•Gravity only affects the vertical

Check your understanding

Supposing a snowmobile is equipped with a flare launcher that is capable of launching a flare vertically (relative to the snowmobile). If the snowmobile is in motion and launches the flare and maintains a constant horizontal velocity after the launch, then where will the flare land (neglect air resistance)?

a. in front of the snowmobileb. behind the snowmobilec. in the snowmobile

http://www.physicsclassroom.com/mmedia/vectors/tb.gif

Suppose a rescue airplane drops a relief package while it is moving with a constant horizontal speed at an elevated height. Assuming that air resistance is negligible, where will the relief package land relative to the plane?

a. below the plane and behind it.b. directly below the planec. below the plane and ahead of it

http://www.physicsclassroom.com/mmedia/vectors/pap.gif

Numbers

•So we’ve been working with displacement a lot in terms of vector addition. But velocity is a vector, too

•A boat moves with velocity 4 m/s across a river•The current is moving downstream. What does the

boat’s velocity relative to an observer on the shore look like?

Review:• A projectile is any object upon which the only force is gravity,• Projectiles travel with a parabolic trajectory due to the influence of

gravity,• There are no horizontal forces acting upon projectiles and thus no

horizontal acceleration,• The horizontal velocity of a projectile is constant (never changing in

value),• There is a vertical acceleration caused by gravity; its value is 9.8

m/s/s, down,• The vertical velocity of a projectile changes by 9.8 m/s each second,• The horizontal motion of a projectile is independent of its vertical

motion.

Consider the cannon on the cliff again:

•Assume it fires the cannonball at 20 m/s•Gravity causes the cannonball to accelerate

downwards at 9.8 m/s/s

Important Concept:

•Horizontal velocity stays the same•Vertical velocity changes by 9.8 m/s every second

http://www.physicsclassroom.com/mmedia/vectors/hlp.gif

How about an upward angle?

•75.7 m/s at an angle of 15o

http://www.physicsclassroom.com/mmedia/vectors/nhlp.gif

How to find initial velocity?

Consider a projectile launched with an initial velocity of 50 m/s at an angle of 60 degrees above the horizontal.

Horizontal and Vertical Displacement

Vertical displacement (height) can be calculated using

y = viyt +½ gt2

The horizontal displacement (distance or range) can be calculated with

x = vixt

Example:

Time

tP = viy/g

ttotal = 2 tP

In summary

Sample Problem

•A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

•Horizontal Info: x= 35 m, vix=?, ax= 0 m/s2

•Vertical Info: y =-22 m, viy=0 m/s, ay= -9.8 m/s2

•Use y = viy • t + 0.5 • ay • t2 to solve for time; the time of flight is 2.12 seconds.

•Now use x = vix • t + 0.5 • ax • t2 to solve for vix

•Note that ax is 0 m/s/s so the last term on the right side of the equation cancels. By substituting 35.0 m for x and 2.12 s for t, the vix can be found to be 16.5 m/s.