projectile motion - PowerPoint by HWQZ0H9

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									              Projectile Motion




Do not try this at home!
                         Projectile Motion

•   Figure 4-16 shows a pirate ship,
    moored 560 m from a fort
    defending the harbor entrance of an
    island. The harbor defense cannon,
    located at sea level, has a muzzle
    velocity of 82m/s.
(a) To what angle must the cannon be
    elevated to hit the pirate ship?
(b)What are the times of flight for the
    two elevation angles calculated
    above?
(c) How far the pirate ship be from the
    fort if it is to be beyond range of
    the cannon?
Equations of motion along a straight line with
            constant acceleration


 v  v0 at         (Equation of velocity)




x  x0  v0t  1 at 2   (Equation of position)
               2
(Cont’d)
                                       Projectile Motion




The range R is the horizontal
distance the Projectile has traveled
when it returns to its launch height




                             The horizontal motion and the vertical
                             motion are independent of each other.
       The Horizontal Motion
• Because there is no acceleration in the
  horizontal direction, the horizon component x 0v
  of the projectile’s initial velocity remains
  unchanged throughout the motion, as
  demonstrated in the following figure:
                 The Horizontal Motion



The horizontal displacement x  x0 from an intial position x0 is given by
the equation of position with a  0, which we write as
                        x  x0  v0 xt
Because v0 x  v0 cos 0 , this becoms
                     x  x0  (v0 cos 0 )t
    The Vertical Motion

The position equation becomes
                  1 2
y - y0  v0 y t  gt but
                  2
v0 y  v0 sin  0 therefore,
                          1 2
y - y0  (v0 sin  0 )t  gt , and
                          2
v y  (v0 sin  0 )t  gt
            The Vertical Motion


Other equations for vertical motions are
v y  v0 sin  0  gt
and
v  (v0 sin  0 )  2 g ( y  y0 )
 2
 y
                 2
                       Home work




• A movie stuntman is to run across a rooftop and jump horizontally off
  it, to land on the roof of the next building . Before he attempts the
  jump, he wisely asks you to determine whether it is possible. Can he
  make the jump if his maximum rooftop speed is 4.5m/s?
                         Home work
• A rescue plane is flying at a
  constant elevation of 1200 m
  with a speed of 430km/h toward
  a point directly over a person
  struggling in the water ( see
  Fig.4-14). At what angle of
  sight  should be pilot release
  a rescue capsule is it is to strike
  (very close to) the person in the
  water?

								
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