# projectile motion - PowerPoint by HWQZ0H9

VIEWS: 26 PAGES: 14

• pg 1
```									              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?

```
To top