Motion II

1. Motion II 2 and 3 dimensional motion

2. Components of Motion • Motion in 2 dimensions • X component • Y component • Motion in 3 dimensions • X component • Y component • Z component

3. Motion in x direction is independent of motion in y direction and z direction. Separate set of equations of motion for each direction.

4. Equations of Motion • 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)

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

6. Independence of x, , z motion • 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.

7. Nerf Gun Experiment • 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.

8. 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.

9. 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

10. 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

11. Elevated Nerf Gun • 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

12. 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.

13. 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

14. 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

15. 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!

16. Review • Motion in 2 dimensions • X component • Y component • Motion in 3 dimensions • X component • Y component • Z component

17. Motion in x direction is independent of motion in y direction and z direction. Separate set of equations of motion for each direction.

18. Class Activity • 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!