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Free Fall. Projectile Motion – free fall, but not vertical. Free Fall :. Used to describe the motion of any object that is moving _____________________________ the only force acting is ________________ no _____________________ , which is a good approximation if object moves ____________
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Free Fall: • Used to describe the motion of any object that • is moving _____________________________ • the only force acting is ________________ • no _____________________ , which is a good • approximation if object moves ____________ • motion can be _________________ or in an arc • known as a ____________________ • the results are independent of ___________ • All of the equations of __________________can • used as long as you use: • a = _______ = ___________________= ____________ • = _____________on or near Earth’s surface • for the time the object is in ________________ . freely through a vacuum. gravity air resistance slowly up or down parabola mass kinematics -9.81 m/s2 9.81 m/s2 down -g constant free fall
Free fall applies to an object that is… fired up or down _____ __________: dropped thrown _________ down: ___________ from rest: at an angle fired ______________: fired _______up: horizontally in flight …only for the time while it is ________________. • In all cases: • d is _________________if the object ends up • __________ the point where it started. • 2. d is _________________if the object ends up • __________ the point where it started. • 3. v is positive if object is going ________________ • 4. v is negative if object is going ________________ • 5. a is _________________________ positive above negative below up or right down or left always -9.81 m/s2
______________ motion • A. Dropped Objects. Vertical Ex 1: A ball is dropped. How far will it fall in 3.5 seconds? equation: given: d = vit + ½ at2 a = -9.81 m/s2 d = 0t + ½(-9.81)(3.5)2 vi= 0 d = ½(-9.81)(12.25) t= 3.5 s unknown: d= ? d = -60. m
Ex. Harry Potter falls freely 99 meters from rest. How much time will he be in the air? equation: given: d = vit + ½ at2 a = -9.81 m/s2 -99 = 0t + ½(-9.81)(t)2 vi= 0 d= -99 m -99 = -4.905t2 unknown: t2 = 20.2 t= ? t = 4.5 s
Ex. A dinosaur falls off a cliff. What will be its velocity at the instant it hits ground if it falls for 1.3 seconds? equation: given: a = -9.81 m/s2 vf = vi + at vi= 0 t= 1.3 s vf = 0 + (-9.81)(1.3) unknown: vf= ? vf = -13 m/s A rock that has half the mass of the dinosaur is dropped at the same time. If it falls for the same time, what will its final speed be? Which will hit the ground first? same neither
B. Objects Fired Up or Down. Ex. A ball is tossed up with an initial speed of 24 meters per second. How high up will it go? given: vf equation: a = -9.81 m/s2 vf2 = vi2 + 2ad vi= 24 m/s 0 = 242 + 2(-9.81)d vf= 0 -576 = -19.6d vi unknown: 29.4 m = d d= ? What total distance will it travel before it lands? 58.8 m What will be its resultant displacement when it lands? 0. m
For a ball fired or thrown straight up: • _______ d each second on way up • ______ d each second on way down • tup = _____________ • ttotal = _______ = __________ • vtop =__________ • 6. atop= __________ • 7. speedup = _______________ • If object falls back to its original • height, then: vf=______ v = 0 less more tdown 2tdown 2tup 0 -9.81 m/s2 speeddown vi -vi vf going up coming down
Ex. Mr. Butchko is fired directly up with an initial speed of 55 meters per second. How long will he be in the air? given: equation: a = Δv/t a = -9.81 m/s2 a = (vf – vi)/t vi= 55 m/s vi -9.81 = (-55 – 55)/t vf= -55 m/s unknown: t = (-110)/-9.81 vf t= ? t = 11 s How much time did he spend going up? t = 5.5 s
Ex. A shot put is thrown straight down from a cliff with an initial speed of 15 m/s. How far must it fall before it reaches a speed of 35 m/s? given: equation: a = -9.81 m/s2 vf2 = vi2 + 2ad vi= -15 m/s (-35)2 = (-15)2 + 2(-9.81)d vf= -35 m/s 1225 - 225 = -19.6d 1000 = -19.6d unknown: 1000/(-19.6) = d d= ? -51 m = d
-10 m/s2 C. Graphical analysis: use a ≈ _____________ Ex: ball dropped from rest v (m/s) 3 1 2 t (s) 0 0 0 -10 5 m -5 -10 1 -10 15 m -10 25 m -10 2 -20 -20 35 m -20 3 -45 -10 -30 -30 4 -80 -40 -10 -40
time total d velocity 0 m 0 s 0 m/s 5 m 5 m 1 s -10 m/s 15 m 2 s 20 m -20 m/s 25 m 3 s 45 m -30 m/s See any patterns?
Ball dropped: vectors vs. scalars displacement distance d d ~ t2 t t velocity speed v v ~ t t t acceleration acceleration a a constant t t
Ex: ball thrown straight up with vi = 30 m/s -10 0 30 0 -10 1 -10 2 3 -10 4 -10 5 -10 -10 6
v (m/s) going up 30 20 25 m 10 15 m 5 m t (s) 1 2 4 5 3 6 -10 -20 -30 slope = ______________ throughout
Ex: ball thrown straight up with vi = 30 m/s -10 30 0 0 20 25 -10 1 40 10 -10 2 45 3 0 -10 4 -10 5 -10 -10 6
v (m/s) going up coming down 30 20 25 m 10 15 m 5 m 5 m t (s) 1 2 4 5 3 6 15 m -10 25 m -20 -30 slope = ______________ throughout
Ex: ball thrown straight up with vi = 30 m/s -10 30 0 0 20 25 -10 1 40 10 -10 2 45 3 0 -10 -10 40 4 -10 -20 5 25 -10 -30 -10 0 6
v (m/s) 30 20 25 m 10 15 m 5 m 5 m t (s) 1 2 4 5 3 6 15 m -10 25 m -20 -30 going up coming down positive d negative d top -10 m/s2 slope = ______________ throughout
Going down: Going up: time v time 3 s 0 3 s 0 5 m 2 s 4 s -10 10 15 m 20 -20 1 s 5 s 25 m 30 6 s -30 0 s v time
At what time is the ball at its highest point? t = 3.0 s What are the v and a at that time? v = a = -10 m/s2 0 How do the the last 3 sec of this example compare to the example of a ball dropped from rest? the same What will the graph of speed vs. time look like? 30 20 10 t (s) 1 2 4 5 3 6
v (m/s) Ex. How does the picture change if ball is thrown up a with different initial speed, say vi = 20 m/s? 30 20 10 t (s) 1 2 4 5 3 6 -10 -20 -30
v (m/s) Ex. What if ball is thrown up with an initial speed vi = 10 m/s? 30 20 10 t (s) 1 2 4 5 3 6 -10 -20 -30
v (m/s) Ex. What if thrown down a with speed vi = 10 m/s? 30 20 10 t (s) 1 2 4 5 3 6 -10 -20 -30 Ball continues down until it strikes the ground.
What remains the same in all of these graphs? acceleration = -9.8 m/s2 Open your 3-ring binder to the Worksheet Table of Contents. Record the title of the worksheet: Free Fall WS
displacement: velocity: vf = vi + at d = vit + ½ at2 With vi = 0 and a = -10 With vi = 0 and a = -10 vf = 0 + (-10)t d = 0t + ½ (-10)t2 vf = -10t d = -5t2 For t = 0, 1, 2, …. For t = 0, 1, 2, …. vf = -10t = -10(0) = 0 d = -5t2 = -5(02) = 0 = -10(1) = -10 = -5(12) = -5 = -10(2) = -20 = -5(22) = -20 = -10(3) = -30 = -5(32) = -45