Chapter 2 Pt 2
Falling Objects and Gravitational
acceleration
Falling Objects
• As objects fall near Earth’s surface, their
acceleration is nearly uniform.
• Beware of thinking that heavier objects fall
faster than lighter objects OR that the speed
of an object is proportional to how heavy it is.
• Galileo was the first to analyze motion of a
freely falling body in an “ideal” situation
where air resistance or other friction was not
a factor.
• He derived the distance an object free falls is
proportional to the square of the time it falls.
d
t
2
Acceleration due to gravity, g
• Galileo was sure air acted as a resistance to
falling objects that are light weight or have a
large surface area.
• Using his “stone analogy” (heavy stone from
2.0m vs 0.2 m will drive a stake deeper into the
ground so it must be moving faster upon
striking the stake), he claimed the speed of all
objects increases as they fall.
• At a given location on the Earth and in the
absence of air resistance, all objects fall with
the same constant acceleration, g. (Accel. due
to gravity = 9.80 m/s2 (about 32 ft/s2).
A note regarding air resistance
• In most cases, the effects of air resistance are
small enough to be negligible and we ignore
them for the most part.
• It is important to note that air resistance is
very noticeable even on heavy objects when
the velocity becomes large… (airplanes? Golf
balls? Sky divers?)
Coordinates for Free fall
• It is up to the reader to decide whether y0 is
positive in the upward direction or the
downward direction, but be consistent
throughout the problem.
• It is important to talk through positive and
negative velocity and acceleration and how
that affects the moving object (tossed up or
dropped down) as time goes on.
Falling from a tower example
• Suppose that a ball is dropped from a tower
70.0 m high. How far will it have fallen after
1.00 s, 2.00 s, and 3.00 s? Assume y is positive
downward. Neglect air resistance.
• Solution: a = g = +9.80m/s2. (B/c down is pos.)
• Using equation 2-10b with v0 =0 and y0 =0
then y1 = ½ at2 = ½ (9.80m/s2)(1.00s)2 = 4.90m
and y2 = ½ (9.80m/s2)(2.00s)2 = 19.6 m for 2 s
and y3 = ½ (9.80m/s2)(3.00s)2 = 44.1m for 3s
Ball thrown upward example
• A person throws a ball upward into the air
with an initial velocity of 15.0m/s. Calculate
(a) how high it goes, and (b) how long it is in
the air before it comes back to his hand once
it leaves.
g
v
g
v
Solution
• Choose +y to be in the up direction and –y
downward.
• Then a=g=-9.80m/s2 and as the ball rises its
speed decreases until it reaches the top where
its velocity is 0m/s for an instant.
• To determine maximum height, calculate the
position of the ball when v=0m/s. At t=0s we
have y0 = 0m, v0 = 15m/s, and a=-9.80m/s2
and we want to find y.
• Use v2 = v02 + 2ay and solve for y
Continuing on…
• Y=
v
2
v0
2a
2
2
0 (15.0m / s)
2
2( 9.80m / s )
11.5m
• The ball reaches 11.5m height above the hand
• Now to calculate how long it is in the air: since
y represents position or displacement, not
total distance traveled, at the lowest points
(start and finish), y=0m.
Wrapping it up…
• Y=v0t + ½ at2
• 0=(15.0m/s)t + ½ (-9.80m/s2)t2
This is easily factored taking out 1 t:
(15.0m/s-4.90m/s2t)t=0 and gives us 2 solutions
T = 0 and t = 15.0m / s2 3.06s
4.90m / s
The first solution gives us time at the start and
the second solution gives us time at landing.
Your turn to Practice
• Please do Chapter 2 Review pgs 44-45 #s 33,
34, 35, 36, 37, 38, 51, & 52.