2. NAME OF MEMBERS
GROUP 7
2
Name ID Contribution
Tasnim Hassan Ahona 23-55085-3 Theory
Shahriar Kabir 23-55092-3 Collision
Md. Nafis-ul Hoque 23-55212-3 Apparatus & Procedure
Anim Chandra Debnath 23-55225-3 Data analysis & Results
Abrar Kabir 23-55095-3 Discussion & Resources
3. 3
MOTIVATION
Football can provide an interesting context for understanding projectile motion in physics. When a player kicks a
football, the ball follows a curved path known as a projectile. Concepts like launch angle, initial velocity, and air
resistance come into play. Analyzing the motion of a kicked football can help students grasp physics principles in a
practical and engaging way.
4. 4
*Projectile motion involves objects moving through the air under the
influence of gravity and an initial velocity.
*It is a two-dimensional motion, comprising horizontal and vertical
components.
*Factors considered include the independence of horizontal and vertical
motion, gravity's influence on vertical motion, and constant horizontal
velocity.
*Trajectories of projectiles follow a parabolic path.
*Important parameters include range, time of flight, maximum height,
and launch angle.
*Real-world applications include sports (e.g., javelin throw, tennis),
engineering (e.g., bullet trajectories), and scientific studies.
*Air resistance and other external forces can complicate projectile
motion, deviating from idealized paths.
*Initial velocity y = dot v 0 , and angle = θο.
x - component of initial velocity vec v_{0x} = v_{0x}*i ; and v 0x =v 0 cos θο.
y - component of initial velocity vec v_{0y} = v_{0y}*j ; and oy 0 =v 0 sin θο.
Theory
5. 5
Horizontal Motion
At any time t, the projectile's horizontal displacement x is
given by
x = vο x*t + 1/2 * ax * t ^ 2
Where acceleration along xaxis,
a_{x} = 0 At any time 7, the projectile’s
Horizontal velocity v_{x} = vοx
Using vο x =vο cos θο we can write ,
x=(vο cos θο ) t ………………………………(1) .
Vertical Motion
At any time t, the projectile's vertical displacement y is
given by
y = v_{0y}*t + x/2 * a_{y} * t ^ 2 [where, a_{y} = - g ,and v 0v
=v 0 sin theta 0 ]
y=(v 0 sin θο )t - 1/2 * g * t ^ 2 .......................(2)
At any time 1, the projectile's vertical velocity
Vy =v 0 sin θο - g * t .....................................................(3)
And
v^2 y =(vο sin θο t )^2 - 2gt...........................(4 )
6. 6
Maximum Height (H)
At maximum height v_{y} = 0 Then eq (4) gives the maximum height (y
= H) * as
H= vο ^2 * sin^2 θο /2a ,.............................................(5 )
And eq (3) gives the time to reach the maximum height as
t= vο sin θο/g……………………. (6)
Horizontal Range
Eq (2) gives the time of flight as
T= 2vo sin θο/g ……………………….(7)
And the horizontal range is obtained as (x = R)
R = vο^2 sin2θο / g………………………..(8)
Angle of Launch
Analyzing Eq (5) and Eq(8) we can write the angle of launch from
Measurement of H and R as,
θ= tan-1 4H/R………………………(9)
7. 7
Collision
A collision is an event in which two or more bodies exert forces on each other
within a short time. The collision between a ball and a fixed smooth surface is
presented in the following fig.
A collision is characterized by three quantities: namely momentum, kinetic energy
and impulse. Collisions can either be elastic or inelastic. Analysis of momentum and
kinetic energies before and after collision can reveal whether the event is elastic or
inelastic. And Impulse is the change in momentum component along particular
direction.The speeds before and after the collision are v₁ and vy, respectively.
Analysis of momentum conservation
As momentum is a vector quantity, so the impulse is calculated for both the x-
component and the y-component of motion. For the x-component Impulse.Jx =
Change in momentum along x-direction dp x =pxf -pxi
Impulse.Jy = Change in momentum along y - direction. dpy =p yf -p yi
Analysis of Kinetic Energy conservation
The kinetic energies before and after the collision are given by
K Ei= 1/2 * m * v_{i}^ 2
And K Ef = 1/2 * m * v_{f} ^ 2
For an elastic collision the Kinetic energy is conserved, i.e., K Ef = K Ei
But for an inelastic collision, K Ef should be less than KE,
9. 9
PROCEDURE
Set up the ramp end aligns with
the table surface.
Place recording paper on the floor,
topped with carbon paper to mark
ball bounces.
Keep recording paper stationary
until data collection concludes, but
carbon paper can be lifted to
check collision points.
Use the marble ball to locate
position O on the floor and
measure the distance from O to a
reference point on the recording
paper.
After data collection, relocate the
paper for measuring S_{1} and
S_{2}
The ball is released from a point
near the top of the ramp, ensuring
no spin is imparted onto it,
allowing it to roll down and
bounce on the floor with minimal
spin.
The procedure is repeated 10
times, always releasing the ball
from the same point on the ramp.
Avg of S1 & S2 are obtained by
circulating the most marks on
paper. The radius of the circle
indicates the uncertainty S1 & S2
Heights h and H are measured with
a meter scale as accurately as
possible.
To measure H, an upright white
cardboard with carbon paper on it
is placed midway between points A
and B. The procedure in step VI is
repeated 10 times, allowing the
ball to hit the cardboard each time.
H is measured from the average
position of the marks X.
The mass of the marble is
measured.
10. 10
Height
h
(cm)
Height
H
(cm)
Average
distance,
(cm)
Uncertaint
y in s1
Average
distance,
S2 (cm)
Uncertainty
in s2 (cm)
Mass of
marble m
(gm)
51.2 25 35.7 2 89.8 3.1 5.7
Quantity Straight line motion Projectile (two dimensional) motion
-------------------------------------------------------------------------------
Horizontal component | vertical component
Initial velocity u Vox = Vocosθo Voy=Vosinθo
Acceleration a Ax = 0 Ay = -g
Velocity at any point V=u + at
V^2=u^2+2as
Vx = Vox Vy=Voy – gt
Vy^2=v^2oy-2gy
Distance S= vt
S=ut+1/2 at^2
X = Voxt Y = V0yt-1/2gt^2
Experimental Data
Analysis
11. 11
Quantity Straight line Projectile (two dimensional) motion
-------------------------------------------------------------------------------
Horizontal component | vertical component
Initial velocity u Vox = Vocosθo Voy=Vosinθo
Acceleration a Ax = 0 Ay = -g
Velocity at any point V=u + at
V^2=u^2+2as
Vx = Vox Vy=Voy – gt
Vy^2=v^2oy-2gy
Distance S= vt
S=ut+1/2 at^2
X = Voxt Y = V0yt-1/2gt^2
RESULT:
Some basic quantities related with projectile motion
before collision
12. SN Quantities Before collision
(impact)
After collision
impact
Comment
1 Kinetic energy , KE = ½ mv^2 KE i = 320.81*10^3 KE f =
180.51*10^3
Not Conservative
2 X-component of momentum
Px =mvx
Pix = 630.021 Pfx=682.86 Not Conservative
3 Y-component of momentum
Py =mvy
Piy = 1805.6 Pfy = 1261.98 Not Conservative
SN Quantities Corresponding Equation Values with units Comment
1 Horizontal impulse that the
floor gives to the ball
Jx = Pfx - Pix 52.83 Positive Impulse
2 Vertical impulse that the
floor give to the ball
Jy = Pfy - Piy -543.62 Negative(oposite
direction of
gravity)
Analysis of impulse
Analysis of collison:
13. 13
SN Quantities Corresponding Equation Values with units
5 Range of the projectile after bounce R = s2-s1 54.1 cm
6 Angle of launch of the projectile after
impact on the ground
θf =tan^-1 (4H/R) 61.586*
7 Speed of the projectile after impact on the
ground
Vf = √R*G/sin2θf Vf =251.67cms^-1
316.78 cmc^-1
8 X component of the velocity of the
projectile after impact
Vfx = Vf sin θf 119.8 cms^-1
9 Y component of the velocity of the
projectile after impact
Vfy = Vf sin θf 221.4 cms^-1
Some basic quantities related with projectile motion after collision
14. 14
Discussion:
The change in momentum was analyzed from the values of Jx and Jy, representing the
change in momentum in the horizontal and vertical directions, respectively.The change
in kinetic energy after collision indicates the energy transfer during the interaction
between the ball and the surface.Observing the type of collision and its characteristics
provides insight into the conservation of kinetic energy and momentum principles.
Resources:
• Fundamental of physics (10th Edition): Projectile motion (Chapter 4,page 70-
75),Collision And Impulse (chapter 9, page 266)
• https://www.google.com/search?sca_esv=2802dad21c2aa82f&sxsrf=ACQVn0_s8NX5Sny6BKSnVqBkEND0rzKKhA:1709145063417&q=projectile+motion+marble+ramp+experiment+pic&tbm=isch&s
ource=lnms&sa=X&ved=2ahUKEwjQnP_C1c6EAxUhTWwGHaW_BlcQ0pQJegQIDBAB&biw=1280&bih=661&dpr=1.5#imgrc=aw0bOV8U8VPfcM
• https://www.google.com/search?q=marble+ball&sca_esv=2802dad21c2aa82f&sxsrf=ACQVn09dAVVL3SJACZ1fhvWGqv8Jpi07Rw%3A1709147623357&ei=54XfZaO7FeegnesP2PONoAo&oq=&gs_lp=
Egxnd3Mtd2l6LXNlcnAiACoCCAQyBxAuGOoCGCcyBxAjGOoCGCcyBxAjGOoCGCcyBxAjGOoCGCcyBxAjGOoCGCcyBxAjGOoCGCcyBxAjGOoCGCcyBxAjGOoCGCcyBxAjGOoCGCcyBxAjGOoCGCcyFBAAGIkF
GOMEGOkEGOoCGLQC2AEBMhQQABiABBjjBBjpBBjqAhi0AtgBATIUEAAYgAQY4wQY6QQY6gIYtALYAQEyFhAAGAMYjwEY5QIY6gIYtAIYjAPYAQIyFhAAGAMYjwEY5QIY6gIYtAIYjAPYAQIyFhAAGAMYjwEY
5QIY6gIYtAIYjAPYAQIyFhAAGAMYjwEY5QIY6gIYtAIYjAPYAQIyFhAAGAMYjwEY5QIY6gIYtAIYjAPYAQJI5ghQAFgAcAF4AZABAJgBAKABAKoBALgBAcgBAPgBAZgCAaACBqgCEpgDBroGBggBEAEYAboGB
ggCEAEYC5IHATE&sclient=gws-wiz-serp
• https://www.youtube.com/watch?v=uEnUG_1TYxc
