First, let’s talk about The River Boat… . If the boat has a speed of 10 m/s, crosses theriver perpendicular to the current and the current is 5 m/s, what is the resultant velocity of the boat? How long to travel across the river?
If the boat has a speed of 10 m/s and the current is 5 m/s, what is the resultant velocity of the boat? VR 5 m/s downstream Θ 10 m/s across VR = √10 2 + 52 = 11.18 m/s Θ = Tan-1 (5/10) = 26.57º downstream
How long to travel across the 120 m wide river? The time to cross depends on the speed across the river. t = d v = 120 m 10m/s = 12 sec How far downstream will the boat land on the far bank? The distance downstream depends on the downstream current speed and the time in the water. d = vt = (5 m/s)(12sec) = 60 m downstream
The BIG Rule; The perpendicular components of motion are INDEPENDENT of each other So… the velocity across the river is independent of the velocity down the river. We will use this rule again and again…
Projectile Motion Thank you Physics Classroom: http://www.glenbrook.k12.il.us/gbssci/Phys/Class/usage.html
What forces are working on the arrow as it flies horizontally through the air? 15 mph
FORCE • A Push or Pull • If velocity constant, the force of thrust is equal but opposite the force of air friction • Is the arrow falling? The downward force working on the arrow is GRAVITY. This is greater than the upward force of air resistance. • Anything thrown or launched on this planet is under the influence of gravity.
What keeps the arrow moving forward? • Inertia • a property of matter that opposes any change in its state of motion • Newton’s First Law
Projectile An object propelled through the air, especially one thrown as a weapon
Projectile Motion • The process of movement horizontally and vertically simultaneously.
Types of Projectile Motion . http://www.glenbrook.k12.il.us/gbssci/Phys/Class/usage.html
Projectiles Facts • Projectiles maintain a constant horizontal velocity (neglecting air resistance) due to 1st law of motion Vi and Vf are equal. We will refer to these as VX (horizontal velocity) • Projectiles always experience a constant vertical acceleration of “g” or 9.80 m/s2 (neglecting air resistance) due to 2nd law of motion • Horizontal and vertical motion are completely INDEPENDENT of each other.
Two Components of Projectile Motion • Horizontal Motion • Vertical Motion • THEY ARE INDEPENDENT OF ONE ANOTHER!!!!!!!!
How would you describe the trajectory? Parabolic
Suppose you shoot a gun a drop a spare bullet at the same time. • Who lands first?
Projectiles. From Physclips: Mechanics with animations and film. • View the independence of vertical and horizontal motion • Ballistics cart demo
Show Mythbusters gun video here • If time permits
EX 1 A cannon ball is shot from a cannon with a horizontal velocity of 20m/s. What is the vertical and horizontal displacement after 1 second? After 2 seconds ?
Vertical displacement:What do you know? d = vi = vf = a = t = horizontal velocity of 20m/s
Vertical displacement:What do you know? d = vi = 0 m/s vf = a = 9.8 m/s2 t = 1sec Which formula wouldyou use to solve for d? dy = viy t + ½ ay t2 horizontal velocity of 20m/s
To calculate vertical displacementONLY USE VERTICAL INFO ! dy = viy t + ½ ay t2 Whatisviy t = to? dy = ½ ay t2 Where: dy = vertical displacement (y axis) ay= g = gravity (9.8m/s2) (sometexts use negative to indicatedownward. Wewill assume gravity to be positive.) t = time in seconds
Horizontal displacement:What do you know? d = vi = vf = a = t = horizontal velocity of 20m/s = vx
Horizontal displacement:What do you know? d = vi = 20 m/s vf = 20 m/s a = 0 m/s t = Wewill use 1s and 2 sec Which formula wouldyou use to solve for d? dx = vix t + ½ ax t2 horizontal velocity of 20m/s = vx
Of these three equations, the top equation is the most commonly used. The other two equations are seldom (if ever) used. An application of projectile concepts to each of these equations would also lead one to conclude that any term with ax in it would cancel out of the equation since ax = 0 m/s/s.
To calculate horizontal displacement. ONLY USE HORIZONTAL INFO ! Time determined vertically. dx = vi t + ½ a t2 Since a is zero, then ½ a t2 = zero dx = vix* t d = vt Where: dx = horizontal displacement (x axis) The subscript x refers to horizontal Vix = initial horizontal velocity t = time in seconds
How does vertical displacement change as time increases? How does horizontal displacement change as time increases?
EX 2 • A ball is thrown horizontally at 25 m/s off a roof 15 m high. • A. How long is this ball in flight? • B. How far does the ball travel vertically? • C. How far does the ball travel horizontally? • How would I calculate final velocity horizontal? Vertical? • Hint: Determine how long the ball is in the air using vertical information, then use calculated time to determine horizontal distance.
How long is it in the air? d = vit + .5at2 Since vi= 0, this can be simplified to: d = .5at2 To solve for t: t = d/.5a 1.75 sec
Using time from vertical motion, can calculate distance for horizontal motion dx = vi t + ½ a t2 Since a is zero, then ½ a t2 = zero dx = vix* t d = vt 43.8m
2 Objects are dropped from a height of 10 m. Object A has a mass of 50 g. Object B has a mass of 100g. If there is no air friction, then: • A. Object A should hit the ground before Object B • B. Object B should hit the ground before Object A • C. Object A and Object B should hit the ground at the same time.
d = vt d / t = v 20 m / 1.01 s = v 19.8 m/s = v d = .5 at2 1.01 s = t
EX 4A projectile is shot with a speed of 39.5 m/s straight off a roof and lands 198 m away. From what elevation was it shot? ? With what speed does it impact the ground vertically and horizontally? With what overall velocity does it impact the ground? d = vt d / v = t 198 m / 39.5 m/s = t 5.01 s = t d = ½ at 2 d = ½(9.8 m/s2)(5.01)2 d = 123 m
EX 4A projectile is shot with a speed of 39.5 m/s straight off a roof and lands 198 m away. From what elevation was it shot? ? With what speed does it impact the ground vertically and horizontally? With what overall velocity does it impact the ground? vxf = 39.5 m/s vyf = at vyf = (9.8 m/s2)(5.01) vyf = 49.1 m/s vr= √(49.1 m/s)2 + (39.5 m/s)2 vr= 63.0 m/s
EX 5 A projectile is shot horizontally off a 267-m tall building with a speed of 14.3 m/s. A. With what speed does it impact the ground vertically and horizontally? B. With what overall velocity does it impact the ground?
vf horizontal is constant at 14.3 m/s • vf2 = vi2 + 2ax to determine vf vertically • vfy = 72.3 m/s • overall velocity? • This is just determining the resultant using Pythagoreans • vr2 = (14.3 m/s)2 + (72.3m/s)2 • vr = 73.7 m/s
Supposing a snowmobile is equipped with a flare launcher which is capable of launching a sphere 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 snowmobile b. behind the snowmobile c. in the snowmobile
Many would insist that there is a horizontal force acting upon the ball since it has a horizontal motion. This is simply not the case. The horizontal motion of the ball is the result of its own inertia. When projected from the truck, the ball already possessed a horizontal motion, and thus will maintain this state of horizontal motion unless acted upon by a horizontal force. An object in motion will continue in motion with the same speed and in the same direction ... (Newton's first law). Remind yourself continuously: forces do not cause motion; rather, forces cause accelerations
Ex. 6 A plane flying at 115 m/s drops a package from 600m. How far from the drop point will it land? Objects dropped from a moving vehicle have the same velocity as the moving vehicle.
Horizontal: Vx = 115 m/s dx = ? Vertical: Viy = 0 dy = 600. m a = 9.8 m/s2 This is the same problem we’ve been working… dy = ½ at2 dx = (115 m/s)(11.1s) 600. m= ½ (9.8m/s2)t2 dx = 1280 m t = 11.1 s
Example 7: A projectile is thrown upward at a rate of 13.22 m/s and at an angle of 83.1° with the horizontal. A. How long is the projectile in the air? B. Calculate the range. C. What is the peak height?