# Projectile Motion Physics 12. Motion in 2D  We are now going to investigate projectile motion where an object is free to move in both the x and y direction.

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Projectile Motion Physics 12

Motion in 2D  We are now going to investigate projectile motion where an object is free to move in both the x and y direction

What is projectile motion?  Any object given an initial thrust and then allowed to soar through the air under the force of gravity only is called a projectile.

Projectile Motion  We know that an object (in the absence of air resistance) that is launched at a given angle should follow a parabolic path

Projectile Motion – Horizontal Launch  An object that is launched horizontally will have no initial velocity in the y direction so the entire initial velocity will be in the x direction  At this point, we are able to treat the projectile using our two equations of motion

Parts of a Projectile Path  Horizontal Distance = range  Height of projectile = altitude, peak

2D Motion of a Horizontal Thrust  Gravity ONLY affects the vertical distance travelled  Gravity is the ONLY force affecting the object (neglect air resistance)  So a x = a y =  v iy = 0 as there is no initial thrust given

Projectile Motion - Equations

Projectile Motion

Projectile Motion Problem  A cannonball is fired horizontally from the top of a 50.m high cliff with an initial speed of 30.m/s. Ignoring air resistance, determine the following: How long it takes to strike the ground How far from the base of the cliff it strikes the ground How fast it is travelling when it strikes the ground

Projectile Motion Problem  Start with y position equation (4)  Sub in known information (h=50.m) and solve for time

Projectile Motion Problem  Now use x position equation (2)  Sub in time and known information (t=3.2s, v ox =30.m/s) and solve for d x

Projectile Motion Problem  Finally we will use equations (1) and (3)  Sub in time and solve for velocity

Projectile Motion Problem  Now, we employ trigonometry and Pythagorean Theorem to solve for the final velocity vxvx vyvy vrvr

Example 2  You throw a rock off a 291m high cliff horizontally at 12.8 m/s.  A) If the river below is 68.5 m wide, will the rock make it across the river?  (98.6m so it will make it across)  B) With what velocity will the rock hit the water/ground?  (76.6 m/s [80.4’])

Projectile Motion  http://videolectures.net/mit801f99_le win_lec04/ http://videolectures.net/mit801f99_le win_lec04/  Page 536-7, questions 1 to 8

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