Trajectory and Projectile Motion. 10-15 minutes to work on finishing the trajectory problems (then go over a few of them). Trajectory and Projectile Motion.
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Describes the motion of an object in TWO dimensions
Neglect air resistance to make it simpler
The ball is in free fall vertically and moves at constant speed horizontally!!!
Gun Pointing Up
Gun Pointing Down
Gun Pointing Up
Gun Pointing Down
y height of victim / wound (m)
y0 height of shooter’s location (m)
v0 initial speed of the bullet (m/s)
θ angle of firing
t time for bullet to travel (s)
g acceleration due to gravity (always 9.8m/s2)
Gun Pointing Up
Gun Pointing Down
y height of victim / wound
y0 height of shooter’s location
v0 initial speed of the bullet
θ angle of firing
t time for bullet to travel
g acceleration due to gravity (always 9.8m/s2)
How height is affect by gravity
How height is affect by vertical velocity
Height of Shooter
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 – 1.225
1.2 = y0 – 17.725
18.925m = y0
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
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– 7.056
125.056 =0.849v0
147.3m/s =v0
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)
-23.751 =-0.1v0sin(59) = -0.1x0.857v0
277.1m/s=v0