Projectile Motion

Projectiles Projectile – an object that is “projected” or shot into the air; once air born it has no means of altering its path. As such a projectile is an object upon which the only force acting on it is gravity. Types of projectiles - Cannon balls, spears, arrows are all considered to be projectiles. Click on the picture to go to the PhET site and open the Projectile Motion SIM. Play with the SIM for a while.

Examples of Projectiles Non-Projectiles Thrown rock Rocket Fired bullet Missile Dropped bomb Powered plane Arrow Bird (unless it is dead!)

Path of Projectile Inertia: an object is in constant motion unless acted on by another force.

Path of Projectile Path with no gravity! With air friction If friction is ignored (i.e. no air, or projectile shot on the moon) then:   Three Big Rules for Projectile Motion: 1.    The only force acting on a projectile is gravity! 2.    Therefore horizontal velocity will remain constant. 3. The vertical velocity is affected by the acceleration of gravity.

Path of Projectile Horizontal Motion Vertical Motion Accele-ration No Yes, “a” is downward at 9.8 m/s2 Velocity Constant Changing by 9.8 m/s every second Since we have objects going in two directions; 1) along the x axis and 2) along the y axis we have to divide information as horizontal motion (ax , vx) and vertical motion (ay , vy)

Simple Projectile To fully analyze projectiles you need to understand vectors. In this course we are going to examine one type of simple projectiles that requires very little knowledge of vectors. The more complicated projectile motion will be covered in Physics 12 Our projectiles will be projected horizontally off a cliff: v = 20 m/s h= 34 m

The motion of this horizontally projected cannon ball The motion of this horizontally projected cannon ball. It has an initial horizontal velocity, vx = 20 m/s. 1) What do you notice about the horizontal velocity of the projectile all through out it’s motion? It stays the same  2) Now examine its vertical velocity, vy. What do we notice about the vertical velocity? It increases speed at 9.8 m/s every second in the negative direction.

After watching the cannon ball’s projections off the cliff, the following data can be surmised. Time Horizontal Velocity Vertical Velocity 0 s 20 m/s, right 1 s 9.8 m/s, down 2 s 19.6 m/s, down 3 s 29.4 m/s, down 4 s 39.2 m/s, down 5 s 49.0 m/s, down NOTICE THE INITIAL VERTICAL VELOCITY….. VERY IMPORTANT

In fact the vx remains constant while the vy increase (downwards) due to the acceleration of gravity. These two facts will allow us to determine the range of any horizontally shot projectile. Since the horizontal velocity remains constant (in the absence of air resistance) then the RANGE (distance from start to final impact) of a projectile, dx, can easily be calculated by: dx = (vx )( t)

If we are given the horizontal velocity and we know the height of the cliff then we can calculate the range of the projectile, only if we know the time it takes for the projectile to land on the ground Out problem is how do we solve for time? We must first determine how long it takes for the ball to drop. Remember the only acceleration on the projectile is the acceleration due to the gravitational force, g = 9.80 m/s2. Therefore a dropped ball will “fall” at the same rate as a horizontally projected ball! v = 20m/s h = 34 m dx = ( vx )(t)

Take two cannon balls, drop one from the top of the cliff as another one is shot horizontally off the same cliff. Note how the “dropped” ball and the projected ball travel the same vertical distance in the diagram shown below: The time for both of the balls to fall to the bottom of the cliff is exactly the same! Of course the projected ball will land down range, but they will both hit the ground at the same time! Don’t believe me! Then check out this Mythbusters clip! We have dealt with falling objects before and therefore should be able to determine the time for the fall of the vertical ball (and therefore the projected ball) from a simple kinematic equation.

Example #1 2 dy t = a (2)( -34) t = -9.80 t = 2.63 s A cannon is shot at 20 m/s off a 34 m high cliff determine the time it takes to hit the ground and the range. We can use the equation dy = viyt + ½ at2 Since the initial velocity of the vertical drop is 0 m/s we can deduce the equation to  dy = ½ at2 It can be rewritten as: a t = 2 dy -9.80 t = (2)( -34) t = 2.63 s

dx = (vx)( t) dx = 20 m/s x 2.63 s dx = 52.6 m = 53 m Since we now know both the horizontal velocity and the time we can determine the Range of this projectile: dx = (vx)( t) dx = 20 m/s x 2.63 s dx = 52.6 m = 53 m

Best Practices to solve Questions 1) Draw out the situation 2) Divide you page into horizontal information and vertical information 3) Identify your known and unknown values 4) Solve

Example #2 Horizontal (x) Vertical (y) dix = ??? diy = -67 m 1. An arrow is shot horizontally off a 67 m high cliff with a velocity of 82 m/s. Determine the range of this projectile. (Your answer should be 303 m) Step 1 - Draw it out Step 2 and 3 - Divide info into Horizontal and Vertical. Then identify known and unknowns Vix = 82 m/s h = 67 m Horizontal (x) Vertical (y) dix = ??? diy = -67 m vix = 82 m/s viy = 0 m/s ax = 0 m/s2 ay = -9.8 m/s2 t = ??

Step 4 – Solve question Part 2 – dx = vix t dx = (82m/s)(3.6977s) dx = 303.211 m = 3.0 x 102 m Part 1 – find the time dy = (viy)(t) + ½ at2 -67 m = (0 m/s)(t) + ½ (-9.8 m/s2)t2 -67 m = 0 – 4.9 m/s2 (t2) −67𝑚 −4.9 𝑚 𝑠 /𝑠 = t2 −67𝑚 −4.9 𝑚 𝑠 /𝑠 = 𝑡 2 t = 3.6977….s t = 3.7 s

2. A blue ball is shot horizontally of a cliff at 23 m/s 2. A blue ball is shot horizontally of a cliff at 23 m/s. It lands at a range of 132 m away from the base of the cliff. What is the height of the cliff? (ans. = 161 m) 3. A red ball is shot horizontally of a cliff and takes 4.5 s to hit the ground. It lands at a range of 132 m away from the base of the cliff. What is the height of the cliff? (ans. = -99 m) What was the horizontal velocity of the ball? (ans. = 29 m/s) 4. A green ball is projected horizontally at 45 m/s off a 27 m high cliff at the same time as a brown ball is dropped from the same cliff. Which ball will hit the ground first? (ans. = at same time) What is the vertical impact velocity of each ball with the ground? (ans. = -23 m/s)