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Motion II

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Motion II

2 and 3 dimensional motion

- Motion in 2 dimensions
- X component
- Y component

- Motion in 3 dimensions
- X component
- Y component
- Z component

Motion in x direction is independent of motion in y direction and z direction.

Separate set of equations of motion for each direction.

- ax= Fx/ m
- vx = vox + axt
- x = xo+ voxt + (1/2)axt2
- vx2 = vox2 + 2ax(x – xo)
- ay= Fy/ m
- vy = voy + ayt
- y = yo+ voyt + (1/2)ayt2
- vy2 = voy2 + 2ay(y – yo)

- az= Fz/ m
- vz = voz + axzt
- z = zo+ vozt+ (1/2)azt2
- vz2 = voz2 + 2az(z – zo)

- Motion in the x direction is independent of motion in the y or z directions.
- Motion in the y direction is independent of motion in the x or z directions
- Motion in the z direction is independent of motion in the x or y directions.

- In class, a nerf gun was fired horizontally from a height of 3’10” and struck the ground at a distance of 16’10”.
- Calculate the muzzle velocity of the projectile.
- Calculate the time of flight of the projectile.

- Neglecting aerodynamic drag, the projectile leaves the muzzle with a velocity vo = vox.
- The projectile as it leaves the muzzle has no velocity in the y-direction, i.e. voy = 0.
- The only force on the projectile after it leaves the muzzle is the force of gravity.
- The acceleration in the y direction (up and down) is g = 32.2 ft/s2.

- Knowing the initial y component of velocity is 0, the acceleration in the y direction is 32.2 ft/sec2, and the distance to the floor is 3’10”,
- 3’10” = 3.833 ft = y – yo
- ay = 32.2
- y = yo+ voyt +0.5ayt2
- 3.833 = 0.5 x 32.2 x t2
- t = 0.488 sec

- In that time of 0.488 sec, the projectile travels
- 16.833 ft in the horizontal direction.
- vx = 16.833/0.488 = 34.49 ft/sec
- So the muzzle velocity is 34.49 ft/sec

- Consider the same nerf gun, but now elevated at an angle of ϴ⁰ to the horizontal.
- The muzzle velocity is vo
- The horizontal velocity vox = vocosϴ
- The vertical velocity is voy= vosinϴ
- vyvo
- vx

- The only force acting on the projectile after it leaves the muzzle is gravity – in the y-direction.
- The projectile will arc up, stop rising, and arc down to hit the ground.
- We can then calculate how high the projectile will rise and the time it takes to reach that maximum height.

- vy2 = voy2 + 2ay(y – yo)
- voy = vosinϴ
- ay = - g
- vy2 = vo2sin2ϴ + (2)(-32.2)(y – yo)
- If ϴ = 30⁰ and vo = 28.0 ft/sec
- 0 = (28.0)2 (0.5)2 – 64.4 (y – yo)
- (y – yo) = 3.04 ft

- To calculate the time for the velocity in the y-direction to go from (14) ft/sec to 0,
- vy = voy + ayt
- 0 = 14 – (32.2)t
- t = 0.435 sec
- The projectile then begins to fall and it takes another 0.435 sec for it to hit the ground. A total time of flight of (2)(0.435) = 0.87 sec

- During that entire 0.87 sec, the projectile is moving in the x-direction at its initial speed.
- There is no force in the x-direction causing it to speed up or slow down.
- Its speed in the x- direction is (28)(0.866) ft/sec = 24.25 ft6/sec
- In 0.87 sec, the projectile travels (0.87)(24.25)
- = 21.1 feet in the x-direction before it impacts the ground!

- Motion in 2 dimensions
- X component
- Y component

- Motion in 3 dimensions
- X component
- Y component
- Z component

Motion in x direction is independent of motion in y direction and z direction.

Separate set of equations of motion for each direction.

- Consider a rifle with a muzzle velocity of 3,000 ft/sec firing at ϴ⁰ to the horizontal.
- Calculate the range and time to impact as a function of ϴ.
- Create an excel worksheet and plot range vsϴ.
- At what value of ϴ would you get the maximum range? Analytically and graphically!