a gravity g falling objects accelerate down at 32 ft/sec 2 9.8 m/sec 2 more precisely: at sea...
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agravity gFalling objects accelerate down at
32 ft/sec2
9.8 m/sec2
more precisely:at sea level 9.80621 m/sec2
32.1725 ft/sec2
16.0 km (10 miles) above earth’s surface agravity drops to about 9.75 m/sec2
32.5 km (20 miles) above earth’s surface agravity drops to about 9.70 m/sec2
even skydivers experience a 9.8 m/sec2
even commercial jet carriers experience agravity only 1% under the value at sea level!
North Pole agravity = 9.832San Francisco agravity = 9.800Denver agravity = 9.796
The gravitational force on an object decreases by about a millionth for every
3 meter (~10 feet) gain in elevation.
An individual with a 50 kilogram mass weighs 490 Newtons (110 pounds) in
New York City; but ~0.25 newton (1 ounce) less in mile-high Denver.
If you drop an object (assuming air resistance is negligible) it accelerates down at g=9.8 m/sec2. If instead you throw it downward, its acceleration after release is
A. <gB. =gC. >g
If you drop an object (assuming air resistance is negligible) it accelerates down at g=9.8 m/sec2. If instead you throw it upward, its acceleration themoment after you release it is
A. <gB. =gC. >g
A ball is dropped from rest, and a bullet shot out of a gun, straight down. Neglecting air resistance, which has the greater acceleration just before hitting the ground?
A) the ball
B) the bullet
C) both have the same acceleration
The acceleration of gravity does not depend on the mass or the speed of the object in free fall!
Inclined Plane
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A ball rolling down an inclined plane has constant acceleration
Is the acceleration of the ball down the ramp 9.8 m/s2?
A) yes B) no
No! This ball is not in free fall. Gravity alone does not act on it.
The inclined plane provides a forceof support which affects the motion!
A ball is thrown straight up and falls back to the ground. Which of the following is true about its velocity v and its acceleration a at the highest point in its path?
A) v = 0 and a = 0
B) v 0, but a = 0
C) v = 0, but a 0
D) v 0 and a 0
E) cannot be determined
At the highest point in its path, the ball momentarily comes to a stop, and so its velocity is zero.
However, since the ball is in free fall, its acceleration is g = 9.8 m/s2 (at every moment).
The maximum velocity, v, an object reaches falling freely from rest, is
directly proportional to the time, t, of its fall: v t .
A) TRUE B) FALSE
How fast is an object moving at the endof a one second fall?
9.8m
sec2 1 sec = 9.8 m/sec
How fast is an object moving at the endof a three second fall?
9.8m
sec2 3 sec = 29.4 m/sec
A ball is in free fall for 8 seconds.Its speed after 4 seconds is half thespeed it will reach by 8 seconds.
A) TRUE B) FALSE
The distance it travels in the first4 seconds equals the distance itwill travel in the last 4 seconds.
A) TRUE B) FALSE
For objects in freefall, the distance fallen, d, is directly proportional tothe time, t, spent falling: d t.
A) TRUE B) FALSE
How far does an object in freefall dropin one second? In 8 seconds?
distance = rate timeYour grade
school mnenomic
We qualify this slightly with
= starting point + rate timecurrent position
tvxx 0
position at t = 0
Since a falling object’s velocity is constantly increasing, maybe we should use:
rate time = (average velocity) time
average velocity = vmin + vmax
2
= v0 + v
2starting velocity
at time=0current velocity
at the present time
tvxxavg
0
position at t = 0
average velocity = v0 + v
2starting velocity
at time=0current velocity
at the present timebuilt up by accelerating
over the time t
tatvv
xx
2
)(00
0
atvv 0
tatv
xx
2
20
0
2
2
1
00attvxx
How far does an object in freefall dropin one second? In 8 seconds?
2
2
1
00attvxx
distance fallen2
2
1
00attvxx
from rest0
22
2
12
2
1 )1(/8.9 ssmatd
= 4.9 m
In 1 second:
22
2
1 )8(/8.9 ssmd
= 313.6 m
In 8 seconds:
150 100 50
Last time we saw:
25
100
225
1 sec3 sec
2 sec1.5 sec
3 sec
1 sec
downhill: vavg=25 cm/sec
final speed: v =150 cm/3sec
downhill: vavg=50 cm/sec
final speed: v =100 cm/sec
downhill: vavg=225 cm/3sec
final speed: v =150 cm/sec
A rocket test projectile is launched skyward at 88 m/sec (198 mi/hr). How high does it go?
vo = ?vfinal = ?
a = ?
time to reach peak, t = ?height reached, x = ?
When does it peak? What has happened by that point?
v = 0 tsec
m
sec
m )8.9(88 2
tsec
m
sec
m )8.9(88 2
tsec
msecm
28.9
88= 9 sec
So 2
2
1
00attvxx )81)(8.9()9)(88( 2
21
2 secsecsec
m
sec
m
= 792m – 396.9m= 395 m (1/4 mile)
9.8 m/sec2
8 m/sec
A ball is thrown straight upward and caught when it returns to the height from which it was released.
1. At its peak position, the ball’sA. instantaneous velocity is maximum.B. instantaneous velocity is zero.C. instantaneous acceleration is zero.D. both B & C are true.
2. The time to fall back from its peak position is A. greater than B. equal to C. less than
the time it took to climb that high.
3. The speed it builds up to downward by the moment it is caught is A. greater than B. equal to C. less than
the speed it was thrown upward with.
For objects for which air friction is negligible,
time up = time downspeed down = speed up
Two spheres of identical mass areReleased when the mechanism aboveIs triggered. One sphere is launched Horizontally by a spring, the other is simultaneously dropped from rest.
A. The launched sphere will reach the ground first.B. Both spheres touch ground at the same time. C. The dropped sphere reaches the ground first.
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15
25
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45
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40
50
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Galileo’s critics argued that if the earth moved:
A stone is flung horizontally at 8 m/secfrom a point 1100 meters above the baseof a cliff.How far away from the cliff does it land?
?
?
/80.9
11002
x
t
smga
my or +1100m
+9.80m/s
Careful!!!Actually ay = 9.8m/s2
ax = ?0
0
/8
0
0
y
x
v
smv
VERTICALLY2
221 )80.9(01100 t
s
mm t = 15 sec
HORIZONTALLY
)15)(/8( ssmx x = 120 m
Its y-component of motionis like dropping from rest!
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25
35
45
10
20
30
40
50
5 10 15 20 25 30 35
Weight Support (floor)
Adding all these supporting forces together
some pull left,some pull right,some pull forward,some pull back
all tend to pull UP!
(a tug-of-war balancing)
Styrofoam bridge
weighted at center
Pressure applied to rigid glass bar
Launched vertically, when the spring is released, from this fast moving cart,the ball will
A. still be caught by the barrel from which it was fired.B. land out in front of the cart since it carried the cart’s forward motion before it was fired.C. land behind the cart which will have moved out from beneath it.