Lab # 111
“Projectile Motion”

Objectives :

The purpose of this lab is to understand projectile motion in two dimensions and apply linear motion equations to solve the problem. Also to predict where a given size bullet to be placed in horizontal direction to leaving table. In addition to that measure the range of projectile leaving table at an angle, and use the value to predict where the projectile bullet to strike at floor.

Theory :
To predict where bullet will land on the floor when lunched from the mini launcher from a table at an angle, but to do that first we need to determine the velocity of the bullet and to determine by launching the bullet horizontally from the table and measuring the vertical and horizontal distances through distance that the bullet travels. Then the initial velocity can be used to calculate where the bullet will land when the bullet lunched from the mini launcher at an angle.
Horizontally lunched bullet at θ=0 off a table with an initial speed, vo, the horizontal distance travelled by the bullet and the vertical distance of the bullet drop in time t describe: x= v0t (1) y=v0yt+12gt2 (2)
Now we can determine the initial velocity of the bullet by measuring x and y. where y = 0, therefore, the time of the bullet can be found by equation: t= 2yg, Where g = 9.8 (3)
Now, to guess the range of the bullet, x, at and angle, θ, above horizontal, we guess the time of the bullet using vertical motion equation:
12gt2-(v0sinθ)t-H=0 (4) and we solve for t , then we use the distance formula to calculate D, D=(v0cosθ)t. (5)
Preliminary experiment : Trial number | 1 | 2 | 3 | 4 | 5 | Average | X cm | 149 | 149.2 | 148.8 | 149 | 149 | 149 |

Measurement of initial velocity : x | y | t= 2yg | v0=xt | 149 cm | 99 cm | 4.495 s | 33.14 |

Part 1 : Calculation and verification y | y0=y-H | t= 2yg | Calculated D=v0t | Measured D | % Difference | 99 cm | 54 cm | 3.31 s | 110 cm | 112 cm | 1.8% | 99 cm | 48.5 cm | 3.14 s | 104.3 cm | 105.3 cm | 1% |

Part 2 : Calculation and verification
At angle θ=300 θ | R | H | v0=Rgsin2θ | t Solving by12gt2-(v0sinθ)t-H=0 | CalculatedD=(v0cosθ)t | Measured D | % Difference | 300 | 0.938 m | 0.09 m | 3.26 | 0.38 s | 1.072 m | 1.081 m | 0.83% |

Discussion:
Q.1 How close your prediction with experiment result? What could be the factors that cause the error? How to improve it?
Ans. Our prediction was almost nearest with experiment results. Theses can have many factors causes the error such as air resistance, firing point, or a surface. Changing an angle and block some resistance can improve the result.
Q.2 What shape is the path of a projectile?
Ans. The shape of a projectile is “PARABOLA.”
Conclusion :
In conclusion, the experiment helped us to understand the projectile motion in two dimensions. We learn that horizontal and vertical motions are two independent of each other expect for their time.