Download and S y

Copyright Sautter 2003
Motion in Two Dimension - Projectiles
• Projectile motion involves object that move up or down and
right or left simultaneously. A ball thrown into the air at an
angle is a common example as is water sprayed from a garden
hose or a bullet fired at a target.
• Any object once released from its accelerating force is
accelerated downward by gravity at all times during its flight.
This causes a continual variation in its vertical velocity while its
horizontal velocity remains unaffected.
• It discussing projectile motion, the horizontal and vertical
components are treated independently. The combined effect of
these two motions on the object give its parabolic flight path.
When objects are projected from a specific horizontal elevation
and land at the same horizontal elevation, the parabolic path of
the object is symmetrical.
• The next slide shows a projectile path and key points during the
flight in terms of displacements(Sy and Sx), velocities (Vyand Vx)
and accelerations (Ay and Ax) ( g, of course, refers to gravity)
CLICK
HERE
JUST AFTER
FIRING
Sy =0, Sx = 0
Vy = +max
Vx = constant
Ay = g, Ax =0
AT HIGH POINT
Sy = max, Sx = ½ max
Vy = 0, Vx = constant
Ay = g, Ax =0
JUST BEFORE
LANDING
Sy = 0, Sx = max
Vy = -max
Vx = constant
Ay = g, Ax =0
A
Y COMPONENT
X COMPONENT
X COMPONENT
Y COMPONENT
C
X COMPONENT
B
Y COMPONENT
Vy = 0
Sy
Vy inst = slope of tangents
Time
Ay inst = slope of tangents
Vy
Ay
Time
Time
Ay inst = - 9.8 meters / sec2
Path of object
Without gravity
Object
Projected at
Angle 
Vo
Actual projectile
path

Distance fallen due
to gravity (1/2 g t2)
Vertical height if gravity did not
act on the object (Vo sin  t)
Sy = actual height
of object
Sx
Sy = height without effects of gravity – distance fallen due to gravity
Sy = Vo sin  t + ½ g t2 (the value of gravity is negative)
Horizontal distance traveled by projectile Sx = Vo cos  t
Water spraying from
a hose is a common
example of projectile
motion
The vertical motion of
the water is accelerated
by gravity. The horizontal
motion is constant
velocity and is unaffected
by gravity!
Vy = Vo sin  + gt
Vx = Vo cos 
Projectile Vx > 0
Constant velocity
Dropped
Object
Vx = 0
Accelerated by
gravity
Vertical displacements are equal for projectiles and dropped objects
however horizontal displacements are greater for projectiles.
Vertical acceleration for both dropped objects and projectiles is that
of gravity (-32 ft/s2, -9.8 m/s2). They both hit the ground at the exact
same time but, of course, the projectile is further away !
Vy = 0
Vy = + max
Vy = - max
Horizontal component of projectile motion
(Constant velocity – no acceleration)
Vertical component of projectile motion
(Accelerated by gravity)
Vertical
Displacement


Horizontal
Displacement
Vertical
Velocity
Horizontal
Velocity
Vertical
Acceleration
Horizontal
Acceleration



Maximum Height
hmax
Maximum Horizontal
Distance (Range)
range

When an object is dropped
at the exact same time a
projectile is fired at the
falling object, aiming
directly at the object
always insures a direct hit.
Why ?
Because the object and the
projectile once fired,
are in both in free fall !
Solving Projectile Problems
A ball is thrown horizontally at 12 m/s from a building 30 meters
high. (a) How long will is be in the air? (b) How far from the base
of the building will it land?
Vo = 12 m/s
•


“horizontally” means the angle of projection is 0 degrees
In part (a) we are asked to find time when the
vertical distance is – 30 meters (negative means
30 m
below the point of release).
In part (b) we are asked to find horizontal distance
(a) Using the equation shown and inserting –30 for
vertical distance, 12 for the original velocity, 0 degrees
for the angle and – 9.8 for gravity (MKS) we get:
-30 = 12 (sin 0)t + ½ (-9.8)t2, solving for t gives 2.47sec
(b) Using the equation shown and inserting 12 for
the original velocity, 0 degrees for the angle and
2.4 seconds for time found in part (a) we get
Vx = 12 (cos 0) 2.47 = 29.6 meters
Solving Projectile Problems
A ball is thrown upward at a 300 angle, at 12 ft/s from a building
300 feet high. (a)How long will is be in the air? (b) How far from
the base of the building will it land?
Vo = 12 m/s
•
300
300 m
•
The object is projected at + 300. Vo = 12 ft/s
and we are asked in part (a) to find the time
when Sy = - 300 ft.
In part (b) we are asked to find the horizontal
distance (Sx).
(a) using the equation shown we insert –300 for Sy, 12
for Vo 300 for the angle and – 32 ft/s2 for gravity (ENG)
and get –300 = 12(sin 300 ) t + ½ (-32) t2 which gives
–300 = 6 t – 16 t2 or 16 t2 –6 t - 300 = 0 (a quadratic).
Using a = +16, b = - 6 and c = - 300 we insert these
values in the quadratic equation find t = -3.21 or + 5.46
The negative t means the ball hits the ground before it
is thrown in is therefore obviously wrong. t = +5.46 sec


(b) Using the equation shown and the time value from
part (a) we get Vx = 12 (cos 300) 5.46 = 56.7 feet
Solving Projectile Problems
A projectile is shot on level ground at a 45 degree angle with a velocity of 20
ft/s. (a) How high will it travel? (b) How far will it go?
hmax
R


• Vo = 20, the angle = 450 and in part
(a) we are asked to find hmax
• In part (b) we are asked to find the
range ( R)
(a) using the equation shown we insert 20 for Vo 450 for
the angle and – 32 ft/s2 for gravity (ENG) and get hmax =
- ((20)2 (sin 450)2 )/ (2 (-32)) which gives (400 x (0.707)2) /
-64 . Maximum height = 3.13 feet.
(b) Using the equation shown we get R = - ((20)2 x (sin
2 x 450)) / - 32 = 12.5 feet
Solving Projectile Problems
A ball is throw downward at an angle of 300 with a velocity of 10 m/s from a
building 40 meters high. What is the velocity of the ball when it hits the ground?
-300
Vo = -10 m/s
40 m



Downward = - 300
We are asked to find the velocity (a vector
quantity - direction counts) when the object
hits the ground. At this point it is moving
both horizontally and vertically.
We will first find the Vx value and the Vy value
and add them using vectors.
Using the given equations, Vo = 10 m/s, g = - -9.8 m/s2
(MKS) and Sy = - 40 m we must first find time (t).
-40 = 10 sin (-300) t + (-9.8) t2 , 9.8 t2 + 5 t – 40 = 0.
Solving the quadratic formula with a = 9.8, b = 5 and c
= -40 we get t = - 2.29 or + 1.78 seconds. Of course, the +
1.78 is the correct answer.
To find the vertical velocity, Vy = 10 (sin – 300) + (-9.8) 1.78,
Vy = -22.5 m/s
To find the horizontal velocity, Vx = 10 (cos –300) 1.78 =
+15.4 m/s
Previous Problem (continued)
• When the object hits the ground it is moving both vertically
under the influence of gravity and horizontally at constant
velocity (unaccelerated).
Vx = +15.4 m/s
Vy = -22.5 m/s
Vresulting
• To find V resulting we use vector addition (the Pythagorean Theorem),
Vr = ((-22.5)2 + (15.4)2)1/2 , Vr = 27.3 m/s
• Vector values such as velocity requires a direction also. Using the
inverse tangent we get, tan –1 (15.4 / 22.5 ) = 34.4 0
• The velocity at which the object hits the ground is 27.3 m/s and an
angle of 34.4 degrees below the horizontal.
Solving Projectile Problems
• A projectile is shot on level ground at a 45 degree angle
with a velocity of 20 ft/s. When will its velocity be –5 ft/s?
Vo = 20
V = - 5, t = ?
450

• Vy = -5 ft/s, Vo = 20 ft/s and the angle of projection = 450.
Using the given equation, -5 = 20 (sin 450) + (- 32 ) t
• t = 0.598 seconds
Check - At the highest point, Vy =0 and 0 = 20 (sin 450) + (-32) t
= 0.442 seconds at the highest point and when descending t must
be larger than 0.442 and smaller than 0.884 (2 x 0.442), the time
when the object lands. The answer is between these two values!
A ball is thrown horizontally at 12 m/s.
(a) What is its velocity after 2.0 seconds?
(A) 12 m/s (B) 19.6 m/s (C) 23 m/s (D) 31.6 m/s
Hint: use
The vector
Sum of Vx + Vy
A ball is thrown at 400 above the horizontal at 4.0 m/s.
What is its horizontal velocity after 0.50 seconds?
(A) 2.6 m/s (B) 3.1 m/s (C) 3.4 m/s (D) 5.5 m/s
What is the vertical velocity in the previous problem Gravity is its
acceleration
After 0.50 seconds ?
It is -9.8 m /s2
(A) 2.6 m/s (B) 3.1 m/s (C) 4.9 m/s (D) 7.5 m/s
(MKS units)
Hint: maximum
What must be the original velocity of a projectile
horizontal
in order to reach a target 90 km away ?
distance occurs
(A) 320 m/s (B) 939 m/s (C) 882 m/s (D) none
at an angle of 450
A ball is thrown at a 300 angle at 10 m/s. How far
on the horizontal will it land ?
(A) 9.8 m (B) 0.88 ft (C) 8.8 m (D) 10.2 m
Click
Here
For
answers
Real Projectile Motion
• In actual free fall situations objects reach a terminal velocity and
do not accelerate at a constant rate throughout the entire fall.
• The buoyant force to the air acts to reduce the acceleration caused
by gravity. The degree to which this retarding force acts, depends
on several variables.
• As the density of the air increases, buoyancy increases. Also,
increased surface area increases the effects of buoyancy and
causes a decrease in terminal velocity.
• The shape and aerodynamic properties of the object also effect
terminal velocity.
• Most importantly, as the velocity of the falling body increases, the
force opposing the fall increases, thus at a specific velocity the
force of gravity and opposing force become equal and terminal
velocity is reached.