Name: Sharon Mohammed
Candidate Number:
School: Iere High School
Class: Form 5s (Science)
Project Title: Projectile Motion
Additional Mathematics SBA:
Teacher: Ms. Lia Chunilal
Date:

TABLE OF CONTENTS
Project Title……………………………………………………………………………pg3

Aim of Project………………………………………………………………………..pg4

Problem Statement……………………………………………………………………pg5

Information about Volleyball…………………………………………………………pg9

Apparatus and Materials………………………………………………………………pg6

Method for Experiment……………………………………………………………….pg7

Solution to Problem…………………………………………………………………..pg8

Verification of Solution……………………………………………………………….pg18

Discussion……………………………………………………………………………..pg19

Conclusion…………………………………………………………………………….pg21

Bibliography...................................................................................................................pg22

PROJECT TITLE:
To determine the ideal angle at which Penelope, a national volleyball player, needs to serve the ball in order to obtain the distance that it needs to go over the net but stay in the court y applying the theory of projectile motion.

AIM:
To determine the ideal angle and distance needed for a volleyball ball to be served over the net within the dimension of the court using projectile motion.

PROBLEM STATEMENT:
Penelope is a national volleyball player and she wants to know the angle at which she needs to serve the volleyball in order to get the ball over the net but also in the court every time she serves. In order to do so, she sets up a projectile launcher and she shot a ball (projectile) at different angles and found the angle at which the ball went a midway distance. From this data, she also found the projectile’s distance for the ideal launching angle.

INFORMATION ABOUT VOLLEYBALL
The overall measurement for a volleyball court is 60 feet by 30 feet. Each side of the court is therefore 30 feet by 30 feet in size. A center line is marked at the center of the court dividing it equally into 30 feet squares. The net is hung directly above the center line at 7 feet 4 inches for women and 8 feet for men. In this case it’s for women.

APPARATUS AND MATERIALS: * Projectile launcher * Projectile (steel ball) * Stopwatch * Chalk * Meter rule * Protractor * Measuring tape * Calculator * Camera

METHOD OF EXPERIMENT: 1. First the projectile launcher was set up and clamped unto the edge of a sturdy table. 2. A meter rule was then positioned in such a way that the 0 meter mark was directly beneath the barrel of the launcher 3. The launcher was then adjusted to be at 0 degrees. The projectile was then launched, and the time it took to stay airborne to the time where it first it the ground was highlighted. The distance on the meter rule was read off, and this measurement and corresponding time was recorded 4. Step no. 3 was repeated 2 more times so that an average could be found. 5. Step no. 3 was also repeated at the following degrees, 10, 20, 30, 35, 40, 45, 50, 55, 60, 70, 80, 90. 6. The vertical distance (w), that is from the barrel to the floor, and the horizontal distance (h) travelled by the ball was found. 7. The projectile pathway was then found for the ideal angle from the formula,
Y= (g/2(u) ²) R
Where Y is the vertical distance the ball falls in the given time, and R is the angle. 8. A graph of angle vs. range was plotted to show the pathway of the projectile.

SOLUTION TO PROBLEM:
Results obtained:-
TABLE SHOWING RANGE (DISATANCE TRAVELLED) AND TIME OF FIGHT OF THE PROJECTILE Angle | Range/m ± 0.0005 (Distance Travelled) | Time of Flight/s | | 1 | 2 | 3 | Average | 1 | 2 | 3 | Average | 0 | 1.90 | 1.84 | 1.86 | 1.87 | 0.47 | 0.43 | 0.44 | 0.45 | 10 | 2.79 | 2.77 | 2.76 | 2.77 | 0.43 | 0.46 | 0.44 | 0.44 | 20 | 3.51 | 3.58 | 3.59 | 3.56 | 0.78 | 0.75 | 0.76 | 0.76 | 30 | 4.35 | 4.32 | 4.40 | 4.36 | 0.60 | 0.63 | 0.60 | 0.61 | 35 | 4.66 | 4.67 | 4.64 | 4.67 | 0.72 | 0.75 | 0.72 | 0.73 | 40 | 4.76 | 4.78 | 4.77 | 4.78 | 0.81 | 0.83 | 0.81 | 0.82 | 45 | 4.55 | 4.58 | 4.56 | 4.66 | 1.03 | 1.00 | 1.02 | 1.02 | 50 | 4.11 | 4.14 | 4.12 | 4.34 | 1.03 | 1.02 | 1.09 | 1.05 | 55 | 3.75 | 3.79 | 3.78 | 3.98 | 0.91 | 1.00 | 0.90 | 0.94 | 60 | 3.46 | 3.40 | 3.44 | 3.57 | 1.03 | 1.03 | 1.00 | 1.02 | 70 | 2.78 | 2.74 | 2.75 | 2.76 | 1.18 | 1.11 | 1.12 | 1.14 | 80 | 1.67 | 1.67 | 1.62 | 1.65 | 1.16 | 1.19 | 1.20 | 1.18 | 90 | 0.21 | 0.24 | 0.26 | 0.23 | 0.80 | 0.94 | 0.81 | 0.85 |

MATHEMATICAL METHOD USED: d- vertical distance/displacement t- time a- acceleration s- distance from starting point u- initial velocity v- deceleration * vertical distance d= (g) (t²)/2 = (10) (0.45²)/2 = 1.01 (2dp) * initial velocity u= d/t = 1.01/0.45 = 2.24 (2dp) * Determining the maximum height achieved by the ball v² = u² + 2(a)(s) using v= 0 = 2.24² + 2(-10)(s) 20s = 2.24² s = 5.0176/20 = 0.25m * Determining the time at the maximum height v= u + (a)(t) using v= 0 = 2.24 + (-10)(0.45) = 2.24 + 4.5 = 6.74
10 degree angle * Time to reach maximum height v= u + 2 (a)(t) since u= Vi sinα , where α is the angle of projection
Therefore
0 = 2.24 sin 10 + (2)(-10)t
20t = 2.24 sin 10 t = 2.24 sin 10/20 = 0.02s (2dp) * Maximum height
Using equation :
Y = (g/2(u)²) R = (10/2(2.24)²) 2.77 = 2.76m (2dp)

20 degree angle * Time to reach maximum height v= u + 2 (a)(t) since u= Vi sinα, where α is the angle of projection
Therefore
0 = 2.24 sin 20 + (2)(-10)t
20t = 2.24 sin 20 t = 2.24 sin 20/20 = 0.04s (2dp) * Maximum height
Using equation:
Y= (g/2(u)²) R = (10/2(2.24²)) 3.56 = 3.55m (2dp)

30 degree angle * Time to reach maximum height v= u + 2 (a)(t) since u= Vi sinα, where α is the angle of projection
Therefore
0= 2.24 sin 30 + (2)(-10)
20t = 2.24 sin 30 t = 2.24 sin 30/20 = 0.06s (2dp) * Maximum height
Using equation
Y= (g/2(u)²) R = (10/2(2.24²)) 4.36 = 4.34m (2dp)

35 degree angle * Time to reach maximum height v= u + 2 (a)(t) since u= Vi sinα, where α is the angle of projection
Therefore
0 = 2.24 sin 35 + (2)(-10)t 20t = 2.24 sin 35 t = 2.24 sin 35/20 = 0.07s (2dp) * Maximum height
Using equation:
Y= (g/2(u)²) R = (10/2(2.24²)) 4.67 = 4.65m (2dp)

40 degree angle
Time to reach maximum height v= u + 2 (a)(t) since u= Vi sinα, where α is the angle of projection
Therefore
0 = 2.24 sin 40 + (2)(-10)t
20t = 2.24 sin 40 t = 2.24 sin 40/20 = 0.08s (2dp) * Maximum height
Y = (g/2(u²)) R = (10/2(2.24²)) 4.78 = 4.76m (2dp)

45 degree angle * Time to reach maximum height v= u + 2 (a)(t) since u= Vi sinα, where α is the angle of projection
Therefore
0 = 2.24 sin 45 + (2)(-10)t
20t = 2.24 sin 45 t = 2.24 sin 45/20 = 0.09s (2dp) * Maximum height
Y = (g/2(u²)) R = (10/2(2.24²)) 4.66 = 4.64m (2dp)

50 degree angle * Time to reach maximum height v= u + 2 (a)(t) since u= Vi sinα, where α is the angle of projection
Therefore
0 = 2.24 sin 50 + (2)(-10)t
20t = 2.24 sin 50 t = 2.24 sin 50/20 = 0.10s (2dp) * Maximum height
Y = (g/2(u²)) R = (10/2(2.24²)) 4.34 = 4.32m (2dp)

55 degree angle * Time to reach maximum height v= u + 2 (a)(t) since u= Vi sinα, where α is the angle of projection
Therefore
0 = 2.24 sin 55 + (2)(-10)t
20t = 2.24 sin 55 t = 2.24 sin 55/20 = 0.11s (2dp) * Maximum height
Y = (g/2(u²)) R = (10/2(2.24²)) 3.98 = 3.97m (2dp)

60 degree angle * Time to reach maximum height v= u + 2 (a)(t) since u= Vi sinα, where α is the angle of projection
Therefore
0 = 2.24 sin 60 + (2)(-10)t
20t = 2.24 sin 60 t = 2.24 sin 60/20 = 0.12s (2dp) * Maximum height
Y = (g/2(u²)) R = (10/2(2.24²)) 3.57 = 3.55m (2dp)

70 degree angle * Time to reach maximum height v= u + 2 (a)(t) since u= Vi sinα, where α is the angle of projection
Therefore
0 = 2.24 sin 70 + (2)(-10)t
20t = 2.24 sin 70 t = 2.24 sin 70/20 = 0.14s (2dp) * Maximum height
Y = (g/2(u²)) R = (10/2(2.24²)) 2.76 = 2.75m (2dp)

80 degree angle * Time to reach maximum height v= u + 2 (a)(t) since u= Vi sinα, where α is the angle of projection
Therefore
0 = 2.24 sin 80 + (2)(-10)t
20t = 2.24 sin 80 t = 2.24 sin 80/20 = 0.16s (2dp) * Maximum height
Y = (g/2(u²)) R = (10/2(2.24²)) 1.65 = 1.64m (2dp)

90 degree angle * Time to reach maximum height v= u + 2 (a)(t) since u= Vi sinα, where α is the angle of projection
Therefore
0 = 2.24 sin 90 + (2)(-10)t
20t = 2.24 sin 90 t = 2.24 sin 90/20 = 0.18s (2dp) * Maximum height
Y = (g/2(u²)) R = (10/2(2.24²)) 0.23 = 0.23m (2dp)

Since the 40 degree angle was the greatest maximum height that was achieved, 40 and 45 degrees was the ideal angles of projection for the steel ball (projectile).

VERIFICATION OF SOLUTION
GRAPH SHOWING RELATIONSHIP BETWEEN RANGES FOR THEIR RESPECTIVE ANGLES

The graph above shows how the angle of projection affects the range while the projectile is airborne.
Scale used: X axis- 1cm, 10 units Y axis- 1cm, 1 unit
Axis: Y axis = Range/ ± 0.0005m X axis = Angles
This graph shows that angle 40 degrees gives the largest range that is the highest point on the graph. It also shows that at 45 degrees the ball started going downwards.

DISCUSSION:
A projectile is an object upon which the only force acting is gravity. There are a variety of examples of projectiles. An object dropped from rest is a projectile (provided that the influence of air resistance is negligible). An object that is thrown vertically upward is also a projectile (provided that the influence of air resistance is negligible). And an object which is thrown upward at an angle to the horizontal is also a projectile (provided that the influence of air resistance is negligible). A projectile is any object that once projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity. Projectile motion is the motion of an object whose path is affected by the force of gravity. We are all affected by gravity, but it profoundly alters the motion of objects that are thrown or shot upward. The arching of a thrown ball is caused by gravity, as well as its falling motion in general.
Gravity is a force that acts on objects; it makes objects accelerate "downward". While we do not need to know about forces to analyze projectile motion we do need to know a very important detail: gravity causes objects to accelerate downward at roughly 9.8 m/s² 9.8 m/s² is the generally accepted amount of acceleration that happens, in some areas of Earth it is more or less but for this experiment I used 10.
Projectile motion is mainly dependent on the angle of projection and velocity of the object. For this SBA the problem created was for a national volleyball player Penelope, to determine the ideal angle at which she needs to serve the volleyball in order to get the ball over the net, but also in the court every time she serves. To find this, a projectile angle was used.

From the results, the graph obtained gave the greatest distance and the highest point at which the ball needs to be served, which was 40 degrees, but at 45 degrees the ball went downwards, which meant that the ball needs to be served at 45 degrees in order to get over the net but still stay in the court.
During the experiment, there were some factors that could have caused an error in the results obtained. These factors include friction due to air resistance produced by the ball, and parallax error. A precaution would be to ensure that the projectile was fully clamped on the edge of the table.

CONCLUSION
To conclude, an angle of 40 degrees, with an average speed of 2.24m/s produced the greatest height travelled by the projectile, but at 45 degrees the ball went downwards. Hence, Penelope needs to serve the ball at 45 degrees in order to get the ball over the net but also in the court.

BIBLIOGRAPHY * http://www.physicsclassroom.com * http://www.andrew.cmu.edu * https://www.google.tt/imghp?hl=en&ei=TD7qVMrKOYS4yQT8wID4Ag&ved=0CAMQqi4oAg