Trajectory and Projectile Motion

Projectile Motion (Physics of Bullets)Describes the motion of an object in

TWO dimensionsNeglect air resistance to make it

simpler

Projectile Motion

The ball is in free fall vertically and moves at constant speed horizontally!!!

•Every moving object has horizontal & vertical velocity.

•Vertical velocity is changed by force of gravity.

•Horizontal velocity stays the same.

Projectile Motion•As a cannon is fired upward, the horizontal velocity stays the same.

•Gravity causes the bullet to arc (which creates the parabola)

•The angle of firing affects the path of the bullet over a period of time.

21)sin(

gttvyy

Gun Pointing Up

Gun Pointing Down

21)sin(

gttvyy

y height of victim / wound (m)y0 height of shooter’s location (m)v0 initial speed of the bullet (m/s)θ angle of firingt time for bullet to travel (s)g acceleration due to gravity (always 9.8m/s2)

Gun Pointing Up

Gun Pointing Down

21)sin(

gttvyy

y height of victim / woundy0 height of shooter’s locationv0 initial speed of the bulletθ angle of firingt time for bullet to travelg acceleration due to gravity (always 9.8m/s2)

Gun Pointing Up

Gun Pointing Down

Height of Shooter

How height is affect by vertical velocity

How height is affect by gravity

21)sin(

gttvyy

1. A sharpshooter is located in a tree. The sharpshooter fires down at a 30 degree angle. It takes the bullet 0.5s to hit the victim’s shoulder at 1.2m high. The speed of the bullet was 62m/s. How high up is the sharpshooter?

1.2 = y0 - 62sin(30)x(0.5) – (0.5)(9.8)(0.5)2

1.2 = y0 – 15.5 – 1.2251.2 = y0 – 17.72518.925m = y0

21)sin(

gttvyy

2. A sharpshooter is located in a tree 15m tall. The sharpshooter fires down at a 25 degree angle. It takes the bullet 0.5s to hit the victim’s shoulder (located at 1.5m high). What is the speed of the bullet as it leaves the gun?

1.5 = 15 - v0sin(25)x(0.5) – (0.5)(9.8)(0.5)2

1.5 = 15 – 0.211v0– 1.225-12.275 =-0.211v0

58.2m/s =v0

21)sin(

gttvyy

3. A sharpshooter fires a gun from the ground at a height of 2.0m and angle of 45degrees up. It takes the bullet 1.2s to hit the victim located in a building 120m high. What is the speed of the bullet as it leaves the gun?

120 = 2.0 + v0sin(45)x(1.2) – (0.5)(9.8)(1.2)2

120 = 2.0 + 0.849v0– 7.056125.056 =0.849v0

147.3m/s =v0

21)sin(

gttvyy

4. A sharpshooter is located in a tree 25m tall. The tree is located 15m away from the victim. It takes the bullet 0.1s to hit the victim’s shoulder (located at 1.2m high). What is the speed of the bullet as it leaves the gun?

1.2 = 25 - v0sin(θ)x(0.1) – (0.5)(9.8)(0.1)2

1.2 = 25 – 0.1v0sin(θ)– 0.049-23.751 =-0.1v0sin(θ)

277.1m/s=v0

-23.751 =-0.1v0sin(59) = -0.1x0.857v0

21)sin(

gttvyy

1. Get into your ballistic lab groups.2. Show me the pre-lab (EVERYONE IN THE GROUP) and

get it approved.3. Pick up your supplies (EVERYone MUST WHERE

GOGGLES).4. Collect your data & make observations.5. Cleanup and put all materials away.

*When done, you can work on your projectile motion practice problems!

A. For each trial of data, calculate the initial speed of the gun. Find an average for the initial speed (v0). (Be sure to show your work!)

B. Use the trajectory method to calculate where the bullet should have hit (height-wise) for each trial.

C. What were some observations you made during the trials?D. Did your results match what you had expected? Why/why not?E. What are some factors that could have affected the outcome of your

results? Explain these factors.

I. Background (Background on anatomy of a gun, how a gun fires, what affects a bullet’s path) (5pts)

II. Purpose (2.5pts)III. Experimental Setup (IV, DV, Control, Extraneous Variables)

(2.5pts)IV. Procedures (no I, we, or you.) (5pts)V. Data Table (5pts)VI. Analysis (MUST BE WELL DEVELOPED) (20pts)VII.Conclusion (review purpose of lab, whether results were expected,

what errors occurred, and how to make this lab more accurate) (5pts)

VIII. Attached: all the lab data sheets/prelab papers (5pts)

Trajectory and Projectile Motion

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