Projectile Motion: Vertical Freefall

1
Projectile Motion Vertical Freefall
2
Projectile Motion Vertical FreefallWhat weve
learned so far

• Aristotles beliefs about gravity
• Some objects have natural place in the center of
  the earth so they fall toward it
• The heavier the object, the stronger the
  attraction
• Acceleration of gravity depends on mass
• Galileos beliefs about gravity
• Used thought experiments to refute Aristotles
  views on gravity (cannonballs)
• Acceleration of gravity is independent of mass

3
Projectile Motion Vertical FreefallWhat weve
learned so far

• Without air resistance, all objects fall at the
  same rate
• The acceleration of gravity is -9.8 m/s2
• Clip of the astronaut on the moon showed hammer
  and feather falling at the same rate

4
Projectile Motion Vertical FreefallWhat weve
learned so far

• Look at a vertical ball toss (no horizontal
  motion)
• Slope of position vs time plot gives velocity
  (varies with time)
• Time to rise equals time to fall to initial
  position

5
Projectile Motion Vertical FreefallWhat weve
learned so far

• Slope of velocity vs time plot gives acceleration
  of gravity
• Slope is constant and negative
• Acceleration of gravity is constant and points
  downward
• Acceleration of gravity is -9.8 m/s2

6
Projectile Motion Vertical FreefallWhat weve
learned so far

• Magnitude and direction of velocity varies with
  time
• Velocity of object is zero at the top of path
• At heights h1h2, the magnitude of the velocity
  is the same
• At heights h1h2, the direction of the velocity
  is the opposite
• for h1h2, v1v2

v
7
Projectile Motion Vertical FreefallWhat weve
learned so far

• Acceleration of object is constant
• Even when object is moving up, the acceleration
  points down

a
8
Projectile Motion Vertical FreefallApplying the
Equations of Motion

• Weve already learned the equations of motion
• v2 v1 at
• d2 d1 v1t 1/2at2
• v22 v12 2a(d2-d1)
• We applied these equations to one-dimensional
  motion in the horizontal direction
• We can apply these equations to vertical freefall
  motion, too!
• When we apply these equations, down is negative
  and up is positive.

()
(-)
9
Projectile Motion Vertical FreefallApplying the
Equations of Motion

• What is the velocity of the ball when it returns
  to a height of 1 meter?
• b. What is the velocity of the ball at the peak
  of its path?
• c. What is the acceleration of the ball at the
  peak of its path?

Vertical Freefall Example 1 John is at bat and
pops the ball straight up. The ball leaves the
bat at a height of 1 meter and with a speed of 90
m/s.
-90 m/s
0 m/s
-9.8 m/s2
10
Projectile Motion Vertical FreefallApplying the
Equations of Motion
Vertical Freefall Example 1, continued John is
at bat and pops the ball straight up. The ball
leaves the bat at a height of 1 meter and with a
speed of 90 m/s. What is the maximum height of
the ball?
Solution 0 (90 m/s)2 2(-9.8 m/s2)(d2 - 1
m) -8100 m2/s2 -19.6 m/s2(d2 -1 m) 413.3 m d2
1 m d2 414.3 m The baseball reaches a
maximum height of 414.3 meters.
Useful equation v22 v12 2a(d2-d1)
11
Projectile Motion Vertical FreefallApplying the
Equations of Motion
Vertical Freefall Example 1, alternate
solution John is at bat and pops the ball
straight up. The ball leaves the bat at a height
of 1 meter and with a speed of 90 m/s. What is
the maximum height of the ball?
Solution 0 90 m/s (-9.8 m/s2)t t 9.2
s d2 1 m (90 m/s)(9.2 s)
(1/2)(-9.8 m/s2)(9.2 s)2 d2 414.3
m Same solution as the first method!
Useful equations v2 v1 at d2 d1 v1t
1/2at2
12
Projectile Motion Vertical FreefallApplying the
Equations of Motion
Vertical Freefall Example 2 Rebecca drops a
water balloon out of a window that is 15 meters
high. Kevin stands directly below the window and
is 1.5 meters tall. How fast is the water
balloon traveling when it hits Kevin?
Solution v22 2(-9.8 m/s2)(1.5 m - 15 m) v22
264.6 m2/s2 v 16.3 m/s The balloon is
traveling at 16.3 m/s when it hits Kevin!
Useful equation v22 v12 2a(d2-d1)
13
Projectile Motion Vertical FreefallApplying the
Equations of Motion
Vertical Freefall Example 3 Catch-up
Problem A boy shoots a rock with an initial
velocity of 21 m/s straight up from his
slingshot. He quickly reloads and shoots another
rock with the same speed 3.0 seconds later. At
what time do the rocks meet?

• Solution
• Rock 1
• d2 d1 v1t 1/2at2
• d2 (21)(t) 1/2(-9.8)t2
• d2 (21)t (4.9)t2
• Rock 2
• d2 d1 v1t 1/2at2
• d2 (21)(t 3) 1/2(-9.8)(t 3)2
• d2 (21)t 63 (4.9)t2 (29.4)t 44.1
• d2 (-4.9)t2 (50.4)t 107.1

14
Projectile Motion Vertical FreefallApplying the
Equations of Motion
Vertical Freefall Example 3 Catch-up
Problem A boy shoots a rock with an initial
velocity of 21 m/s straight up from his
slingshot. He quickly reloads and shoots another
rock with the same speed 3.0 seconds later. At
what time do the rocks meet?
Solution Set distances equal to each
other. (21)t (4.9)t2 (-4.9)t2 (50.4)t
107.1 107.1 (29.4)t t 3.64 seconds The rocks
meet 3.64 seconds after the first rock is shot.
The rocks meet 0.64 seconds after the second rock
is shot.
15
Projectile Motion Vertical FreefallApplying the
Equations of Motion
Vertical Freefall Example 4 Try One on Your
Own! Mary leans over the side of a bridge which
is 35 meters high. She tosses a rock straight up
in the air with an initial velocity of 10 m/s.
How long does it take the rock to strike the
ground below?
Solution v22 (10 m/s)2 2(-9.8m/s2)(0 m 35
m) v2 28.0 m/s v2 -28.0 m/s -28 m/s 10
m/s (-9.8 m/s2)(t) t 3.9 s
Useful equations v22 v12 2a(d2-d1) v2 v1
at
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Projectile Motion
Horizontally Launched Projectiles
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Horizontally Launched ProjectilesWhat weve
learned so far

• Horizontally launched projectiles represent a
  combination of two types of motion
• Vertical motion with constant acceleration
  (gravity)
• Horizontal motion at a constant velocity
• We have already used the equations of motion on
  both types of situations!

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Horizontally Launched ProjectilesCannonball
Example Reviewed

• Horizontal portion of motion
• Imagine a horizontally launched cannonball moving
  with gravity switched off
• Cannonball moves at a constant velocity in a
  straight path (no forces, no acceleration)

v
Gravity off
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Horizontally Launched ProjectilesCannonball
Example Reviewed

• Vertical portion of motion
• Imagine cannonball dropped off of edge of cliff
  with gravity switched back on
• Cannonball moves like objects in vertical
  freefall (constant force of gravity, constant
  acceleration)

Gravity off
v
v
Gravity on
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Horizontally Launched ProjectilesCannonball
Example Reviewed

• Horizontally launched projectile is a combination
  of the gravity-free horizontal path and the
  vertical freefall path
• Resulting path is parabolic

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Horizontally Launched ProjectilesCannonball
Example Reviewed
Yes. The force of gravity acts downward.
None
Yes. Acceleration due to gravity is downward
(9.8 m/s2).
None
Constant
Changing
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Horizontally Launched ProjectilesApplying the
Equations of Motion

• Weve already learned the equations of motion
• v2 v1 at
• d2 d1 v1t 1/2at2
• v22 v12 2a(d2-d1)
• We applied these equations to one-dimensional
  motion in the horizontal direction
• We also applied these equations to vertical
  freefall motion
• We can apply these equations to the horizontal
  motion of HLPs and to the vertical motion of HLPs
  independently!

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Horizontally Launched ProjectilesApplying the
Equations of Motion
Horizontally Launched Projectiles Example 1 A
pool ball leaves a 0.60-meter high table with an
initial horizontal velocity of 2.4 m/s. How long
does is take the pool ball to hit the ground?
Notes for filling out table -Because there is
no acceleration in the horizontal direction, the
final velocity is the same as the initial
velocity -All of the initial velocity is in the
horizontal direction so the initial vertical
velocity is zero -The times in the horizontal
and vertical columns are equal
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Horizontally Launched ProjectilesApplying the
Equations of Motion
Horizontally Launched Projectiles Example 1 A
pool ball leaves a 0.60-meter high table with an
initial horizontal velocity of 2.4 m/s. How long
does is take the pool ball to hit the ground?
Solution The time it takes for the pool ball to
hit the ground is the same as if it were a ball
in vertical freefall. (Remember the simultaneous
ball drop demo?)
Vertical d2 d1 v1t 1/2at2 0 m 0.60 m
½(-9.8 m/s2)t2 t 0.35 seconds
Useful equations d2 d1 v1t 1/2at2
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Horizontally Launched ProjectilesApplying the
Equations of Motion
Horizontally Launched Projectiles Example 1
continued A pool ball leaves a 0.60-meter high
table with an initial horizontal velocity of 2.4
m/s. How far does the pool ball land from the
base of the table
Solution This is simply the horizontal distance
that the pool ball could travel in the time that
it took to fall.
0.84 m
Horizontal d2 d1 v1t 1/2at2 d2 (2.4
m/s)(0.35 s) d2 0.84 meters
Useful equations d2 d1 v1t 1/2at2
26
Horizontally Launched ProjectilesApplying the
Equations of Motion
Horizontally Launched Projectiles Example 2 A
soccer ball is kicked horizontally off a
22.0-meter high hill and lands a distance of 35.0
meters from the edge of the hill. Determine the
initial horizontal velocity of the soccer ball.
Solution Use the vertical motion to determine
the time of flight. Use the time of flight to
determine how fast the ball must have been moving
to travel 35.0 meters in that time.
16.5 m/s
Vertical d2 d1 v1t 1/2at2 0 m 22 m
½(-9.8 m/s2)t2 t 2.12 seconds
Horizontal d2 d1 v1t 1/2at2 35 m v1(2.12
s) v1 16.5 m/s
16.5 m/s
2.12 s
2.12 s
27
Horizontally Launched ProjectilesApplying the
Equations of Motion
Horizontally Launched Projectiles Example 2
continued A soccer ball is kicked horizontally
off a 22.0-meter high hill and lands a distance
of 35.0 meters from the edge of the hill.
Determine the magnitude and direction of the
velocity of the soccer ball when it strikes the
ground.
Solution The final velocity is a combination of
the horizontal velocity and the vertical
velocity. We need to know both!
Vertical v2 v1 at v2 (-9.8 m/s2)(2.12 s) v2
-20.8 m/s
16.5 m/s
-20.8 m/s
16.5 m/s
Final Velocity vt2 vh22 vv22 vt2 (16.5
m/s)2 (-20.8 m/s)2 vt 26.5 m/s
2.12 s
2.12 s
28
Horizontally Launched ProjectilesApplying the
Equations of Motion
Horizontally Launched Projectiles Example 2
continued A soccer ball is kicked horizontally
off a 22.0-meter high hill and lands a distance
of 35.0 meters from the edge of the hill.
Determine the magnitude and direction of the
velocity of the soccer ball when it strikes the
ground.
Solution Use simple trigonometry to find the
direction of the final velocity.
sin(x) 20.8 / 26.5 x 51.7 degrees The soccer
ball hits the ground with a speed of 26.5 m/s and
at an angle of 51.7 degrees to the horizontal.
16.5 m/s
16.5 m/s
-20.8 m/s
16.5 m/s
20.8 m/s
26.5 m/s
2.12 s
2.12 s
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Projectile Motion Review
30
Projectile Motion Review

• Question 1
• The acceleration of gravity
• depends on the mass of the object.
• depends on the initial velocity of the object.
• depends on the distance from the equator.
• is the same for all objects.

31
Projectile Motion Review

• Question 2
• The slope of a position versus time graph gives
• acceleration.
• velocity.
• no useful information.

32
Projectile Motion Review

• Question 3
• The velocity of an object at the peak of its path
• is zero.
• is always zero in the horizontal direction.
• is always zero in the vertical direction.
• is the maximum velocity the object experiences.

33
Projectile Motion Review

• Question 4
• A projectiles acceleration
• is zero at the peak of its path.
• is at a maximum at the peak of its path.
• is positive on the way up and negative on the way
  down.
• is constant.

34
Projectile Motion Review

• Question 5
• John throws a baseball straight up with an
  initial velocity of 15 m/s.
• What is the velocity of the ball at its highest
  point?
• 15 m/s
• -15 m/s
• 0 m/s

35
Projectile Motion Review

• Question 6
• John throws a baseball straight up with an
  initial velocity of 15 m/s.
• What is the acceleration of the ball at its
  highest point?
• 0 m/s2
• 9.8 m/s2
• -9.8 m/s2

36
Projectile Motion Review

• Question 7
• John throws a baseball straight up with an
  initial velocity of 15 m/s.
• What is the velocity of the ball when it returns
  to John?
• 15 m/s
• -15 m/s
• 0 m/s

37
Projectile Motion Review

• Question 8
• A plane wants to drop supplies to the town below.
  As it flies overhead, when should the plane drop
  the supplies such that they reach the targeted
  town?
• before the plane reaches the town
• when the plane is directly above the town
• after the plane passes over the town

38
Projectile Motion Review

• Question 9
• Jill and Bill are traveling at a constant
  velocity on a snowmobile. Jill throws a snowball
  straight up thinking it will land on the people
  riding behind them. Where does the snowball
  really land?
• Jills got it right. Jill and Bill ride right
  out from under it, and it lands on the people
  riding behind them.
• Actually, it lands in front of them. The
  velocity of the snowmobile pushes it forward.
• Haha! It lands right on top of them!