TOPIC 1: VECTORS AND MOTION
. . . continuation
RECAP
• Vectors are quantities that have both magnitude and
direction.
• Vectors have x and y components. The sum of vectors is
called the resultant vector.
• Graphical method and Analytical/Component method
can be used to find the Resultant.
• To describe motion:
– Position () : Displacement ()
: Distance (d)
– Velocity () : Average Velocity ()
: Instantaneous Velocity ()
RECAP
– Acceleration () : Average acceleration ()
: Instantaneous acceleration ()
• Kinematic equations (can only be used if the acceleration
is constant)
GRAVITY AND FALLING BODIES
GRAVITY
- pulls us downward and is
responsible for our weight
- first observed by Isaac
Newton when he was studying
the falling of an apple from an
apple tree
Acceleration due to gravity:
g = - 9.8 m/s2
FREELY FALLING BODIES
Freely falling objects are
objects falling under the
sole influence of gravity.
Do freely falling bodies
encounter air resistance?
No
Are bodies thrown upwards
freely falling bodies?
Yes if there is no influence
of air resistance.
WHY, GALILEO GALILEI?
Under normal conditions In vacuum
“all objects free fall at the same rate of acceleration,
regardless of their mass.”
Where g is equal to -9.8 m/s2.
SYMMETRY IN FREE FALL
• The velocity at the highest point is
ALWAYS zero.
• The time it takes for the a body to go to
its highest point is equal to the time it
takes for the body to go down to its
initial position.
• The initial velocity of the body (launch
velocity) is equal in magnitude to its
final velocity (velocity as it goes back
to its initial position).
SAMPLE PROBLEMS
1. A youngster throws a rock from a bridge into the river
50 m below. The rock has a speed of 15 m/s when it
leaves the youngster’s hand. Calculate the velocity of
the rock when it strikes the water if (a.)it is thrown
straight downward initially, and (b.) it is thrown straight
upward initially
Illustration
Solution
• Given:
(+) thrown straight upward, (-)
thrown straight downward
• Unknown:
• Solution
Use the formula –
So,
Both thrown upward or downward
2. A bullet is launched with an unknown velocity straight
upwards and stays in the air for 60 s.
a. What is the initial velocity of this bullet?
b. What is the maximum height reached by the bullet?
• Given:
• Unknown:
• Solution
a. Use the formula –
Note that at highest point, the bullet stops then back to the
ground. So,
(initial velocity)
• Given:
• Unknown:
• Solution
b. Use the formula –
So,
Motion in 2 Dimensions
Projectile motion
As observed the biker is neither moving in the vertical
or horizontal axis. This is an example of an object
moving in two dimensions.
Projectile
- any object that once projected or dropped
continues in motion by its own inertia and is
influenced only by the downward force of gravity.
Are these projectiles?
YES!
In terms of vector…
IMPORTANT POINTS TO REMEMBER
There is constant horizontal motion which means that the
horizontal velocity DOES NOT CHANGE at any point in the
trajectory.
The vertical motion is identical to that of a freely falling body,
that is, it has constant acceleration, g.
What equations will we use?
• Note that we can solve for the magnitude of v
by using Pythagorean theorem.
Maximum height (Vertical range)
vx v0 cos 0 v y v0 sin 0
At maximum height
• Highest point in the projectile’s trajectory; 𝑣𝑦 = 0
• Now, from equation , we can derive
At maximum height
• Now, from equation
It is dependent to both initial velocity and
angle of the velocity
Greatest value of for the same occurs
when
Horizontal range
Horizontal range
• Horizontal distance from the starting point on the ground until it hits the ground
again later 𝑡.
Example
• A batter hits a baseball so that it leaves the bat
with at an angle of
a. Find the position and velocity of the ball at
t=2.00 s
b. Find the time to reach the maximum height.
c. Find the time to reach the horizontal range
Assume that the ball leaves the bat at t=0.
Assume that the height when the ball leaves the bat is at
Example
Solution
a. Find the position and velocity
of the ball at t=2.00 s
For the velocity:
Solution
a. Find the position and velocity
of the ball at t=2.00 s
For the position:
Solution
b. Find the time to reach the
maximum height.
For the time:
For the maximum height:
Solution
c. Find the time to reach the
horizontal range
Solution
c. Find the time to reach the
horizontal range
For the time:
Recall the Quadratic Equation:
1 2
𝑦 = 𝑦 0 +𝑣 0 𝑦 𝑡 − 𝑔 𝑡
2
𝑦 − 0=(29.6
𝑚
𝑠
)𝑡 −
1
2 ( 𝑚 2
9.8 2 𝑡
𝑠 )
𝑚 𝑚
(
0= 29.6
𝑚
𝑠
𝑡− )
1
2
𝑚 2
(
9.8 2 𝑡 − y
𝑠 ) 𝑎=− 4.9
𝑠
2
; 𝑏=29.6
𝑠
; 𝑐= 𝑦 =44.7 𝑚
Solution
c. Find the time to reach the
horizontal range
For the time:
Solution
c. Find the time to reach the
horizontal range
For the horizontal range position (x):
)
Activity
A man stands on the roof of a 15.0 m tall building and throws a rock with a
velocity of magnitude 30.0 m/s at an angle of 33.0° above the horizontal. You
can ignore air resistance. Calculate:
a. the maximum height above the roof reached by the rock
b. the magnitude of the velocity of the rock just before its strikes the ground.
𝑣𝑜
𝑣𝑜 𝑦
𝛼
𝑣 𝑜𝑥
15 m
End of the
lesson
