Projectile Motion

Who are the MASTERS of projectile motion? ..\..\o Projectile- An object that is thrown or launched by some means through the air Projectile motion- the path a projectile takes as it flies through the air. Who are the MASTERS of projectile motion? ..\..\o

Projectile Motion How many “Things/Forces” are acting on an object launched at an angle ? What is the force? Gravity (down) and… Inertia (horizontal - not a force)

Projectile Motion Horizontal vs Vertical aspect of problems y X The horizontal and vertical aspects of each problem have different vectors acting on them. Thus, they must be treated separately!

Projectile Motion y A second look..\..\ X Describe only the vertical aspect of the flight in regards to velocity and direction (y). Vertically, gravity slows velocity on the way up and increases it on the way down. (in this situation) ***What is the displacement? ZERO! Muy importante- starts and ends on ground

THE horizontal aspect of this flight may be a bit tricky at first… Before we watch the video again, check out the AIR DROP!

Air drop! What do you notice?

Projectile Motion y Recap..\..\ X Describe only the horizontal aspect of the flight is regards to velocity and direction (y). Horizontally speaking, it moves with the same velocity the entire time! Is there displacement horizontally? YES!

Cliff Problems ******Vi = 0!**** Use this aspect for horizontal displacement, velocity, etc. V=d/t used because the horz. (x) Velocity is constant! d=vit+1/2 at2 used because the vert. (y) object will be accelerating due to gravity’s effect. Use this aspect for vertical displacement, velocity, etc. ******Vi = 0!****

A ball is thrown horizontally off of a 44. 0 m cliff. It lands 65 A ball is thrown horizontally off of a 44.0 m cliff. It lands 65.0m from the base of the cliff. How fast was it thrown? 44.0m 65.0m

This problem has both horizontal and vertical aspects to consider. Vertically speaking, the ball does not start falling until it leaves his hand. So Vi =0 Horizontally speaking, it does have a velocity as it leaves his hand! 44.0m 65.0m

65.0m Important time concept! The ball will have the same constant horizontal velocity the entire flight… until it hits the ground. So the horizontal time is linked to the vertical time it takes the ball to fall to the ground. 44.0m 65.0m

Horizontal V? (x direction) We want the horizontal velocity at which he threw the ball. Vi=? Horizontal V? (x direction) dx= 65.0m Vi= ? 44.0m What else do we know horizontally? Nothing! For now this is a dead end… 65.0m

Vertical (y) 65.0m dy= 44.0m Vi=? a= -9.80 m/s2 *Vi= 0! Consider the Vertical aspects (y direction) Vi=? Vertical (y) dy= 44.0m a= -9.80 m/s2 44.0m *Vi= 0! What other vertical data is “known” but not given? 65.0m

65.0m 1) Find out how long it takes the ball to hit the ground (y). A good way to approach this problem… 1) Find out how long it takes the ball to hit the ground (y). 2) Realize that the ball is moving horizontally (x) for that same amount of time 44.0m 3) Use that time in the horizontal aspect 65.0m

Vertical (y) 65.0m dy= 44.0m Vi=? a= -9.80 m/s2 *Vi= 0! d=vit + ½ at2 Consider the Vertical aspects (y direction) Vertical (y) Vi=? dy= 44.0m a= -9.80 m/s2 *Vi= 0! d=vit + ½ at2 t=? 44.0m d= ½ at2 65.0m

65.0m Vi=? d= ½ at2 2d/a =t 2 (44.0 m) =t 9.80 m/s2 Cal= 2.9966596709 Consider the Vertical aspects (y direction) Vi=? d= ½ at2 2d/a =t Setup with units 2 (44.0 m) =t 44.0m 9.80 m/s2 Cal= 2.9966596709 R= 3.00 sec. CAREFUL! 65.0m Use vertical distance!

65.0m 1) Find out how long it takes the ball to hit the ground (y). A good way to approach this problem… 1) Find out how long it takes the ball to hit the ground (y). 2) Realize that the ball is moving horizontally (x) for that same amount of time 44.0m 3) Use that time in the horizontal aspect of the lab 65.0m

Horizontal V? (x direction) We want the horizontal velocity at which he threw the ball. Vi=? Horizontal V? (x direction) t= 3.00 sec dx= 65.0m (don’t use 44.0!) Vi= ? 44.0m V=d/t Why can we use a velocity formula if the ball is accelerating downward? 65.0m

Horizontal V? (x direction) We want the horizontal velocity at which he threw the ball. Vi=? Horizontal V? (x direction) t= 3.00 sec dx= 65.0m (don’t use 44.0!) Vi= ? 44.0m V=d/t Because its horizontal velocity stays the same (is constant) the whole flight! 65.0m

Or about 45mph to the right! Horizontal V? (x direction) t= 3.00 sec dx= 65.0m (don’t use 44.0!) 65.0 m Vi= ? Vi= 44.0m 3.00 sec. V=d/t Cal= 21.66666666667 R= 21.7 m/s 65.0m

Golf Problems Ex. maximum height the ball reaches Use the yellow (x) component for anything that is horizontally related. Use the red (y) component for anything that is vertically related. Ex. how far right it ends up going downfield. Ex. maximum height the ball reaches If you know the diagonal initial velocity… consider its components…

A ball is hit at a 50. 00 with an initial diagonal velocity of 44 A ball is hit at a 50.00 with an initial diagonal velocity of 44.0 m/s. How much time will the ball be in the air? B) How high will it go? C) How far will it go downfield?

First things first… Solve for both components of the initial velocity.

Solve for both components of the initial velocity. Vertical component Sine! Sin θ= opp/hyp 44.0 m/s (33.7 m/s) opp Hyp. Sin 50.00= Opp. 44.0 m/s Cal= 33.7059555 m/s 50.00 Cal= 33.7 m/s Adj.

Solve for both components of the initial velocity. Horizontal component Cosine! Cos θ= adj/hyp 44.0 m/s adj Hyp. Cos 50.00= Opp. 44.0 m/s Cal= 28.28265483 m/s 50.00 Cal= 28.3 m/s (28.3 m/s) Adj.

Which of these velocities stays constant the whole flight? 28.3 m/s = Initial Vx(horz) 44.0 m/s (33.7 m/s) 50.00 (28.3 m/s)

How long is the vertical velocity 33.7 m/s? Only at the starting instant- then it immediately starts to decline (gravity only works vertically). 44.0 m/s (33.7 m/s) 50.00 (28.3 m/s)

How much time will the ball be in the air? Let’s start vertically (horz. Is a dead end for now) Carefully name ALL the VERTICAL knowns…

How much time will the ball be in the air? Be sure all data is strictly vertical! Vert. time (y) Vi= 33.7 m/s a= -9.80 m/s2 *d= 0! (start and ends on ground!) t=?

How much time will the ball be in the air? Be sure all data is strictly vertical! Vert. time (y) Vi= 33.7 m/s a= -9.80 m/s2 *d= 0! (starts and ends on ground!) t=? *d= vit + ½ at2 Insert 0 for d, then solve for t 0= vit + ½ at2

0= vit + ½ at2 -vit -vit -vit = ½at2 t t -Vi= ½ at ½ a ½ a -2Vi = t a

This is the total flight time R= 6.88 sec. Solving for time Vert. time (y) -2Vi Vi= 33.7 m/s = t a a= -9.80 m/s2 *d=0 t=? -2(33.7 m/s) = t -9.80 m/s2 Cal=6.87755102 This is the total flight time R= 6.88 sec.

B) How high will it go (max height)? This is a VERTICAL question. (33.7 m/s) 50.00 (28.3 m/s)

Consider the Whole flight… 1) WHERE is the max height? 2) WHEN is the max height? So in this case 3.44 sec. ½ through the total flight time! Total flight time= 6.88 sec. Max height occurred at… (Keep same number of sig. figs.)

Max height (y) vert. Vi = 33.7 m/s a= -9.80 m/s2 Be sure it’s negative! t= 3.44 sec. (1/2 total flight) dy=? 3.44 sec. dy=?

Max height (y) vert. Vi = 33.7 m/s dy=vit + ½ at2 a= -9.80 m/s2 dy=? Don’t forget dy=(33.7 m/s)(3.44 sec) + ½ (-9.80 m/s2)(3.44sec)2 Cal= 57.94336 R= 57.9 m high 3.44 sec.

C) How far will it go downfield? This is a horizontal question. Remember that the ball moves at a constant velocity in the x direction. So DO NOT use an acceleration formula! Also rem. The time is the same as the WHOLE flight time 6.88 sec. Use V=d/t dx=? (28.3 m/s)

Range downfield (d x) Vi= 28.3 m/s v=d/t t = 6.88 sec d=vt dx=? dx = (28.3 m/s) (6.88 sec.) Cal= 194.7 R= 195 m 3.44 sec. (28.3 m/s)

NOTE! In this problem, the displacement is zero! Vertically speaking, the ball starts and ends up on the ground. Overall the displacement is zero! So d=vit + ½ at2 becomes 0= vit + ½ at2 Now solve for t! You get t= -2vi/a Try it! Golf Problems Horizontally speaking, the time is the same as what you find vertically. Rem. That the horizontal velocity never changes. It’s not accelerating! Use v=d/t or d=vt (when manipulated) Note: the time is the same both horizontally and vertically. The ball is in the air for only 1 amount of time- no matter what direction you consider. So if you can find time vertically, you can use that same time horizontally. This is VERY common in projectile motionn.

Golf Problems For max height calculations: IT is a VETICAL problem. Use the red component. D is NOT zero because at its max height the ball is far from the ground! This max height occurs exactly halfway through the flight. The time it takes to reach this height will therefore be exactly half the time of the entire flight! Golf Problems D= (displacement from ground at max height) Note you put in half the total flight time two times. Don’t forget to square Use d= vit + ½ at2

Conceptual Fun with Projectile Motion!

Like Monkeys?

What would happen… If he lets go the instant you fire (at the monkey)? If there is no Gravity (hanging onto tree)? If he lets go the instant you fire (above the monkey)? What if the gun is fired at a lower velocity?

Classic Physics Question If there is no Gravity?

Here’s a classic Physics Question If he lets go the instant you fire (at the monkey)?

Classic Physics Question If he lets go the instant you fire (above the monkey)?

Classic Physics Question What if the gun is fired at a lower velocity?

Bellwork: A goalie kicks a soccer ball with an initial velocity of 15.5 m/s at an angle of 51.00. How high will the ball go? How far downfield will it go? 7.35 m 23.9 m

Bellwork: A bowling ball is rolled off of a rooftop that is 6.50 m tall. If it lands 8.00 m away from the house, how fast was it going when it first left the roof? T=1.15 sec Vi=6.96 m/s

Launching a projectile… With NO gravity With normal velocities With just the right velocity it “balances gravity”

Projectile motion clips Clip horizontal vs vertical effects a..\..\MaN.VOB Clip horizontal vs vertical effects b..\..\MasEN.VOB Clip no monkey same tricks..\..\demos\Disk 2\02EN.VOB Clip car and marble firing a..\..\Master mon.VOB car and marble firing b..\..\MasbEN.VOB Clip part C .. of physos\Disk 2\03cEN.VOB clip..\..\Masts\Disk 2\05EN.VOB Clip firing at angles for distance a ..\..\MEN.VOB Clip firing at angles for distance b..\VOB