Projectile Motion - Laboratory Report - PHYSICS EXPERIMENT

Experiment No. 3

PROJECTILE MOTION WITH THE USE OF BALLISTIC DEVICE

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I. Objectives The objects of this experiment are as follows: 

By using a projectile launcher, calculate the initial velocity of a ball shot horizontally.



Identify the angle of projection that will generate the maximum rate.



By using a set angle identify the average rate of the projection of the ball.

II. Theory Projectile motion is an example of motion with constant acceleration when air resistance is ignored. An object becomes a projectile at the very instant it is released (fired, kicked) and is influenced only by gravity. The x- and y-components of a projectile’s motion are independent, connected only by time of flight, t. Consider two objects at the same initial elevation. One object is launched at an angle ✓ = 0! at the same moment the second object is dropped. The two objects will land at the same time. This allows the two dimensions to be considered separately. To determine where a projectile will land, one must know the object’s starting position, ⃗⃗⃗ 𝑟0 , initial velocity, ⃗⃗⃗⃗ 𝑣0 , and the acceleration it experiences, ⃗⃗⃗⃗ 𝑎0 . Position as a function of time is then described as: 1

𝑟(𝑡) = ⃗⃗⃗ 𝑟0 + ⃗⃗⃗⃗ 𝑣0 + 3 𝑎𝑥 𝑡 2 Eq. 4.1 is a vector equation; it can be resolved into x and y-components: 1

𝑥 = 𝑥0 + 𝑣0𝑥 𝑡 + 3 𝑎𝑥 𝑡 2 1

y = 𝑦0 + 𝑣0𝑦 𝑡 + 3 𝑎𝑦 𝑡 2 Velocity changes constantly in projectile motion. While horizontal acceleration is zero for the purposes of this experiment, the vertical component of a projectile’s velocity can be described as follows, with ay = g: 𝑣𝑦 = 𝑣0𝑦 + 𝑎0𝑦 𝑡

These are the kinematic equations for constant acceleration. Taken together, they describe the motion of projectiles and other constant-acceleration systems.

III. Materials and Setup     

Ballistic Device Ball Meter Stick Bond paper Marker

Set-up

IV. Methodology

 Calculate the Initial Velocity Procedure 1. Set the projectile launcher at 𝜃 = 0o which can allow us to fire the ball straight out. 2. Use a white sheet of bond paper. Place them on the floor in order to record the landing of the ball. 3. Verify that the ball leave a mark on the white paper after the projectile is fire each time with the use of a marker. 4. The projectile should be fired five times. 5. By using a meter measure the distance from the launcher to where the ball lands, and then calculate the average.

 Identify the Angle of Maximum Range Procedure 1. Fasten the projectile launcher to a stool or any steady platform. 2. One member of the group should set the projectile at 20 degrees angle, and then start adding 10 degrees until the projection angle is 70 degrees. Therefore, the projection angle will go from 20 degrees to 70 degrees. 3. The other member should record the distance travel for the projectile (ball) after it is fired. To do so this member should use a piece of bond paper and meter stick along with a marker. 4. The bond paper should be placed on the floor in order to record where the projectile lands.

 Estimate and Measure the Range at a Particular Angle θ Procedure 1. Predict the velocity, position and time in the vertical direction by using trigonometry and an appropriate kinematics equation. 2. Set you projectile launcher at you chosen angel. 3. Fire the ball five times. 4. Compute the average range. 5. Compared the measured value to the predicted one.

V. Data and Results

Calculating the Initial Velocity at 𝟎𝐨 Trial

x (m)

1

1.19

2

1.22

3

1.19

4

1.17

5

1.20

Δx

1.19 Table 1

Thus, Δx = 1.19 m which is the horizontal distance traveled for the ball, and Δy = 0.56 m is the vertical distance from what the ball falls. To calculate the initial velocity release from the launcher it is necessary to use a kinematic equation. Δy is given by this equation: 1

Δy = 2g𝑡𝑦2 . However, we know t = 1

𝛥𝑥 𝑉𝑥

provided that 𝑡𝑥 = 𝑡𝑦 = 𝑡. Therefore, it is possible to

∆𝑥 2

solve this equation Δy = 2g( 𝑉 2 ) for Vx. 𝑥

Therefore,

1

∆𝑥 2

∆𝑥 2 𝑔

Δy = 2g( 𝑉 2 ) = 𝑉𝑥 = √ 2∆𝑦 𝑥

𝑉𝑥 = √

𝑚 𝑠

(1.19)2 (9.8 2 ) 2(0.56𝑚)

𝑉𝑥 = 3.52

𝑚 𝑠

Identifying the Angle of Maximum Range Angle of Projection (𝛉𝐨 )

Range of Motion (m)

20

1.54

30

2.05

40

2.73

50

2.41

60

2.05

70

1.26

Table 2

y-displacement

3 2 1

x-displacement Graph 1

Estimate and Measure the Range at a Particular Angle θ 

By estimation

By using trigonometry it is possible to determine the components in y and x of the initial velocity. 𝑣𝑥 = 𝑣𝑜𝑥 (𝑐𝑜𝑠𝜃) 𝑣𝑦 = 𝑣𝑜𝑦 (𝑠𝑖𝑛𝜃)

At Ɵ = 40𝑜 𝑚

𝑣𝑥 = (3.52 𝑠 )(cos40) = 𝑚

𝑣𝑦 = (3.52 𝑠 )(sin40) =

Illustration 1

𝑣𝑥 = 2.70 m/s 𝑣𝑦 = 2.26

𝑚 𝑠

The time of flight is given by 1 𝑦 = 𝑦𝑜 + 𝑣𝑦𝑜 𝑡 − 𝑔𝑡 2 2 𝑦 = 0 Because the ball hits the ground at the end. Therefore, we can use this equation and solve for t. At 𝜃 = 40𝑜 𝑚

𝑣𝑦𝑜 = 2. 26 𝑠

𝑦𝑜 = 0.56m 1

0 = 𝑦𝑜 + 𝑣𝑜𝑦 𝑡 − 2 𝑔𝑡 2

0 = (0.56𝑚) + 2.26

Thus, we can use the quadratic formula

𝑚 𝑡 − 4.9𝑡 2 𝑠

−𝑏±√𝑏 2 −4𝑎𝑐 2𝑎

where a = -4.9, b= 2.26, and c = 0.56

𝑡 = 0.98𝑠 ∆𝑥 = 𝑣𝑥𝑜 𝑡 ∆𝑥 = (2.70

𝑚 )(0.98𝑠) 𝑠

∆𝑥 = 2.65 𝑚 

By performing the projectile experiment at an angle θ = 40𝑜 Trial

Range (m)

1

2.75

2

2.71

3

2.73

4

2.72

5

2.73

Average

2.728

VI. Analysis of the Data By estimation at an angle θ = 40𝑜 , we ended with a quadratic equation which describes the path of travel for the object. In this cases it is a parabola. We used this equation to solve for t, and when we take the square root of something we will get two answer one positive and the other negative. The negative time represents the time before the motion starts.

If we compare the measured range to the predicted one, we will get the fallowing percent error. % error = |

(2.65)−(2.728) (2.65)

|x 100 = 2.94%

If the initial velocity is doubled, the range will be this 𝑣= ∆𝑡 =

∆𝑥 ∆𝑡

𝑣0 𝑠𝑖𝑛𝜃 ± √𝑣𝑜2 𝑠𝑖𝑛𝜃 2 + 2𝑔𝑦 𝑔

∆𝑥 = 𝑣0 𝑐𝑜𝑠𝜃 (

𝑣0 𝑠𝑖𝑛𝜃±√𝑣𝑜2 𝑠𝑖𝑛𝜃2 +2𝑔𝑦

∆𝑥 = 𝑣𝑜 𝑐𝑜𝑠𝜃(

𝑔

𝑣0 𝑠𝑖𝑛𝜃 + 𝑣0 𝑠𝑖𝑛𝜃 ) 𝑔

∆𝑥 = 𝑣𝑜 𝑐𝑜𝑠𝜃( ∆𝑥 = ∆𝑥 =

2𝑣0 𝑠𝑖𝑛𝜃 ) 𝑔

2𝑣𝑜 𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃 𝑔

2(2𝑣0 )2 𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃 𝑔

8𝑣𝑜2 𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃 ∆𝑥 = 𝑔 8𝑣2 𝑜 𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃 𝑔 Range = 2 2𝑣𝑜 𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃 𝑔

Range = 4

)

If the lab stool is twice as high, the range will change in this way:

∆𝑥 = 𝑣0 𝑐𝑜𝑠𝜃(

𝑣0 𝑠𝑖𝑛𝜃 ± √𝑣𝑜2 𝑠𝑖𝑛𝜃 2 + 2𝑔(2𝑦) ) 𝑔

∆𝑥 = 𝑣0 𝑐𝑜𝑠𝜃(

0 ± √0 + 2𝑔(2𝑦) ) 𝑔

∆𝑥 = 𝑣0 𝑐𝑜𝑠𝜃√4(𝑦) ∆𝑥 = 𝑣0 𝑐𝑜𝑠𝜃(2)√(𝑦) Range:

2𝑣0 𝑐𝑜𝑠𝜃√(𝑦) 𝑣0 𝑐𝑜𝑠𝜃√2(𝑦) Range:

2 √2

The distance traveled horizontally from the launch position to the landing position is known as the range. The range of an angled-launch projectile depends upon the launch speed (which is constant for this experiment) and the launch angle (the experimental variable). Some sources of error of this experiment was the method used to record the lands of the ball. The paper was not properly attach to the floor, so in this case the distances that we measure from the experiment may vary a little from the calculated ones.

VII. Conclusion On projectile motion, we have learned that objects don't accelerate horizontally but they decelerate vertically. Projectile motion is in action every time an object is thrown, dropped, ejected or shot. To solve projectile motion problems we relate the motion's horizontal component to its vertical component. Most times the time will relate the components together, because the object moves horizontally only as long as it moves vertically. The acceleration in the vertical direction of the projectile is gravity. Gravity decelerates an object at a constant rate, and an object will fall past its takeoff point with the same speed as it took off with. Therefore, the acceleration of the horizontal direction is 0 because the gravity is not acting there.

VIII. Answer to Questions & Problems 1. What is the trajectory of a projectile? Answer In physics, the ballistic trajectory of a projectile is the path that a thrown or launched projectile or missile without propulsion will take under the action of gravity, neglecting all other forces, such as friction from aerodynamic drag. 2. An object is thrown upward at an angle θ above the ground, eventually returning to earth. (a) Is there any place along the trajectory where the velocity and acceleration are perpendicular? (b) Is there any place where the velocity and acceleration are parallel? If so, where? Answer (a) Yes; when the object is at its highest point (b) No. 3. A tennis ball is hit upward into the air and moves along an arc. Neglecting air resistance, where along the arc is the speed of the ball? (a) a minimum and (b) a maximum? Answer (a) When the ball is at its highest point of trajectory (b) At the initial and final positions of the motion 4. A stone is thrown horizontally from the top of a cliff and eventually hits the ground below. A second stone is dropped from rest from the same cliff, falls through the same height, and also hits the ground below. Ignore air resistance. Is each of the following quantities different or the same in the two cases? (a) Displacement (b) Speed just before impact with the ground (c) Time of flight Answer (a) The displacement is greater for the stone thrown horizontally. (b) The impact speed is greater for the stone thrown horizontally. (c) The time of flight is the same for both stones.