DOWNLOAD FREEFALL HANDOUT
Free Fall Motion
• An
object is said to be in the state of free fall if gravity is the only force
acting on it
• Assumption
• The
motion is along the vertical
• Gravity
is the only force
• Acceleration is always 9.8m/s2
• Air
resistance is neglected
• Scenarios
• Dropped
• Viy
= 0
• Acceleration
is 9.8m/s2 as it falls
• Thrown
Downward
• Non
zero initial velocity
• Acceleration
is 9.8m/s2 as it falls
• Thrown
upward
• Non
zero initial velocity
• Slow
down on its way up
• Zero
velocity at the peak
• Accelerate
at 9.8m/s2
Example
• A
ball was dropped from a very high building
– A.
How far below its dropping point will it have a speed of 1960m/s2 .
How many seconds after it was dropped will have this speed? Ans.
– B.
What is the speed of the ball and how far is it below its dropping point 5.0s
after it was dropped?
– C.
How many seconds since the ball was dropped will it be 3,136m below its
dropping point? What will be the velocity at the moment?
– What
is the acceleration at the peak
– F.
What is the acceleration 38.0sec after it was fired
– G.
What is the velocity 50s after it was fired?
– I. At what time is it 3926 m above the ground?
A ball is dropped from rest from the top of a building.
Find:
a) The
instantaneous velocity of the ball after 6 sec.
b) How
the ball fell.
c) The
average velocity up to that point.
PROJECTILE MOTION
-
A projectile motion is a case of free falling
motion where velocity of the object has a horizontal component.
-
It is also a free falling motion and Gravity is
the only force acting on it
-
Has no air resistance
Example
Volleyball,
cannon , baseball
Path of the projectile: Like a parabola
Acceleration
Since there is only acceleration in the vertical direction, the velocity in the horizontal direction is
constant, being equal to Vi cos θ. The vertical
motion of the projectile is the motion of a particle during its free fall. Here
the acceleration is constant, being equal to 
.[1] The components of the acceleration are:
ax = 0
ay = -9.8m/s2 or 32.2
ft/s2
Note: Acceleration is entirely vertical along the
path of projectile
Velocity
The horizontal component
of the velocity of the object remains unchanged throughout the motion. The vertical
component of the velocity increases linearly, because the acceleration due to
gravity is constant. The accelerations in the 
and 
directions can be integrated to solve for the
components of velocity at any time 
,
Note: Velocity is entirely
horizontal at the peak of the projectile
Velocity remain the same at the
entire horizontal.
Initial Velocity (Vi) is
the beginning of the projectile motion,
-
Cannot be zero
Angle (θ) - angle of elevation with respect to
horizontal ground
There are two components
of Velocity
Vix = Vi cos θ (horizontal Component)
Viy = Vi sin θ (Vertical component)
Maximum Height (dymax) – highest altitude the object reach during the
projectile motion
Vy = 0
dymax = Viyt + ½ gt2
dymax = Vfy – ½
gt2
2gdymax = Vfy2
– Viy2
2dymax = (Vfy +
Viy)t
Time to reach the maximum height (t)
t = (Vfy
– Vi y)/g
TOTAL TIME OF FLIGHT (T) =
2 x t
MAXIMUM RANGE – Rmax
Rmax = Vix T
Example:
An arrow leaves
a bow at 30m/s at angle of 30° from the horizontal.
a) What is the max Range (R)
b) Find the maximum Height (dymax)
c) How long it would take the arrow to land on the
other side = (t = time)
Solve first
for the Horizontal and Vertical Component of initial velocity
Viy = ?
Vix =?
Then solve for
Max. Height (dymax) Answer: 11.5m
Then solve for
time at Max height : answer: 1.5s
Since max
height is at halfway of the projectile, therefore
Total time of flight (T) = 2t Answer: 3.0s
Solve for Max
range:
R =
Vix t Answer: 78m
2. Supposed
that a coin rolls off the edge of a table 1.5m high at a horizontal speed of
1.0m/s
a. Find the time
it would take before the coin land in the floor
b. Find the
Maximum range
Since Vi is
already in horizontal, Vix = 1.0m/s
To solve for
the time: use
dymax = Viyt + ½ gt2 Answer: 0.55s
Given:
dymax = -15m (since it is downward)
g = -9.8m/s2
Solve for Maximum Range:
R = Vix t Answer: 0.55m
