Projectile Motion Do not try this at home!. Projectile Motion Figure 4-16 shows a pirate ship,...
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Projectile Motion Do not try this at home!
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Transcript of Projectile Motion Do not try this at home!. Projectile Motion Figure 4-16 shows a pirate ship,...
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
(Equation of velocity )0
v v at
(Equation of position)20 0 1
2x x v t at
(Cont’d)
Projectile Motion
The horizontal motion and the vertical motion are independent of each other.
The range R is the horizontal distance the Projectile has traveled when it returns to its launch height
The Horizontal Motion
• Because there is no acceleration in the horizontal direction, the horizon component
of the projectile’s initial velocity remains unchanged throughout the motion, as demonstrated in the following figure:
0xv
The Horizontal Motion
0 0
0 0
0 0 0
The horizontal displacement from an intial position is given by
the equation of position with , which we write as
Because cos , this beco
0
x
x
x x x
x x v t
v v
a
0 0 0
ms
( cos )x x v t
The Vertical Motion
20 0
0
20 0 0
y 0
0 0
0
The position equation becomes
1- but
2sin therefore,
,an1
- ( sin )2
v ( sin
d
)
y
y
y y v t gt
v t
y y v t gt
v
t
v
g
The Vertical Motion
0
0
2
0
20 0
Other equations for vertical motions are
and
( sin
sin
) 2 ( )
y
y
v v g
y y
t
v v g
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?
