PROJECTILE MOTION by dffhrtcv3

VIEWS: 12 PAGES: 13

									PROJECTILE MOTION

 CH 3 – 2 DIMENSIONAL MOTION
       A gun with a muzzle velocity of 1000 ft/sec
is shot horizontally. At the same time an identical
bullet is dropped alongside the gun. Which bullet
will hit first?
       A gun with a muzzle velocity of 1000 ft/sec
is shot horizontally. At the same time an identical
bullet is dropped alongside the gun. Which bullet
will hit first?




    Guess what! They both land at the SAME time
    but at different places. What’s different?
Projectile Motion
   The difference is that one bullet has
    INITIAL VELOCITY.
Projectile Motion
   The difference is that one bullet has
    INITIAL VELOCITY.
   Projectiles: Objects that are launched or
    thrown into the air and are subject to
    gravity.
Projectile Motion
   The difference is that one bullet has
    INITIAL VELOCITY.
   Projectiles: Objects that are launched or
    thrown into the air and are subject to
    gravity.
   Projectile Motion: Free fall with an initial
    horizontal motion.
Summary of Equations
Projectiles – Horizontal Launch
Vertical          Horizontal
y = ½ gt2         x = vxt

vy,f = gt         vx = constant

vy,f2 = 2gy
Example problem
   A movie director is shooting a scene
    that involves dumping a stunt dummy
    out of an airplane into a pool. The plane
    is 10m above the ground traveling at
    22.5m/s in the positive x direction.
    Where in the path should the dummy
    be dropped in order to land in the pool?
Example cont.
                        Given:

          22.5 m/s      vx = 22.5 m/s

                        y = -10 m
                 10 m
                        g =-9.8 m/s2

                        Find x
Answer
1. Get time t from y = 1/2gt2
      t = √2y/g) = √(2(-10m)/-9.8m/s2)
           = 1.43 sec
2. Find x for 1.43 sec ->x = vxt
                     =22.5m/s *1.43s
                     = 32.2 m
So drop the dummy 32m before the pool.
Instantaneous Velocity
   To find the final velocity of the dummy
    just before it hits the pool, you take the
    vector sum of the x and y components
    of the velocity.
   vx is always a constant 22.5m/s
   vy,f = gt = -9.8m/s2(1.43 sec) = 14m/s
   Use Pythagorean Theorem and tan-1 to
    find the vector sum
Answer
   v = √((22.5m/s)2 + (14m/s)2)
       = 26.5m/s

   θ = tan-1(14/22.5) = 32°
Homework
   p. 102, #1-4 and p. 115, #36

								
To top