Transcript projectile. - Pixelcowboy Physics

Topic 9.2 Space
9.2.2 Projectile Motion
Objectives
 State the independence of the vertical and the
horizontal components of velocity for a projectile in a
uniform field.
 Describe and sketch the trajectory of projectile motion
as parabolic in the absence of air resistance.
 Describe qualitatively the effect of air resistance on the
trajectory of a projectile.
 Solve problems on projectile motion.
Projectile Motion
Components of Motion
 When a body is in free motion, (moving through
the air without any forces apart from gravity
and air resistance), it is called a projectile.
 Normally air resistance is ignored so the only
force acting on the object is the force due to
gravity
 This is a uniform force acting downwards
 Therefore if the motion of the projectile is
resolved into the vertical and horizontal
components
 The horizontal component will be unaffected as
there are no forces acting on it
 The vertical component will be accelerated
downwards by the force due to gravity
 These two components can be considered as
independent factors in the motion of a projectile in a
uniform field
 In the absence of air resistance the path taken by any
projectile is parabolic
Solving Problems
 In solving problems it is necessary to consider the 2
components independently
Comparison of Horizontal and
Vertical Projectile Motion
 Therefore for horizontal motion it is necessary to use
the equation
 speed = distance
time
 Where speed is the horizontal component of the
velocity
 Therefore for vertical motion it is necessary to use the
kinematic equations for uniform acceleration
Example
 A ball is kicked at an angle of 40.0o with a
velocity of 10.0 ms-1. Taking g = 9.81 m.s–2.
How far does it travel horizontally?
10 m.s-1
40o
 To be able to calculate the horizontal distance we
need to know the horizontal speed, and the time.
 The horizontal distance is easy to calculate by
resolving the velocity
10.0 sin 40.0o
10.0 m.s-1
40.0o
10.0 cos 40.0o
 However, to calculate the time we will need to use the
vertical component and the kinematics equations
 s=?
 u = 10.0 sin 40.0o ms-1
 v=?
 a = -9.81 m.s-2 (Up is positive, therefore acceleration
here is negative)
 t=?
 We only have 2 of the values when we need three to
find any other
 However, if we ignore air resistance, then the final
vertical component of the velocity will be equal and
opposite of the initial component
 i.e. v = -10.0 sin 40.0o m.s-1
 Looking at the equations for uniform acceleration, we
need an equation that links u, v, a and t.
 v = u + at
 Rearranging to make t the subject
 t=v–u
a
 Substitute in
 t = -10.0 sin 40.0o – 10.0 sin 40.0o
-9.81
 t = 1.31 seconds
 Now returning to the horizontal components
 Using speed = distance
time
 Rearranging distance = speed x time
 Distance = 10.0 cos 40.0o x 1.31
 Distance = 10.03 = 10.0 metres
Example
 A golf ball is hit horizontally at 25.0 m.s-1 from the top
of a 68.0 m cliff. Sketch the trajectory of the ball. How
long does it take the ball to land? How far from the
base of the cliff does the ball strike the ground? Find
the impact velocity of the ball. [3.72 s, 93.1 m, 44.3
ms-1, 56˚ with the horizontal]
Example
 A golf ball is hit into the air at a velocity of 35.4 m.s-1 at
an angle of 39.0 º to the horizontal. Make a sketch of
the ball’s motion and find the:
 Total time in the air. [4.55 s]
 Time when maximum height is reached. [2.27 s]
 Maximum height. [25.3 m]
 Range (Horizontal distance) [125 m]
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Example 1
M02/430/H(2)
This question is about a projectile.
(a)
!
A girl stands on the edge of a vertical cliff and throws a stone upwards at an angle of 60 to
the vertical such that the stone eventually lands in the sea below. The stone leaves her hand
with a speed of 12 m s -1 at a height of 30.0 m above the sea.
60!
ñ ñ ñ ñ ñ ñ ñ
ñ ñ ñ ñ ñ ñ ñ
30.0 m
12 m s -1
ñ ñ ñ ñ
ñ ñ ñ ñ
30.0 m
Example 1 (cont.)
Taking the acceleration due to gravity to be 10 m s -2 and ignoring air resistance determine
(i)
the maximum height, measured from sea-level, reached by the stone.
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(ii)
the speed with which the stone hits the sea.
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uestion A2 continued)
(b)
M02/430/H(2)
Example 1 (cont.)
In the space provided below sketch, using the same axes, graphs to show how the horizontal
and the vertical components of velocity of the stone vary with time from the moment it
leaves her hand to just before it hits the sea. (Note that this is a sketch graph; you do not
need to add values to the axes.)
[2]
2. (a)
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use v 2 = 2 gh ;
with v = 12 cos 60 = 6.0 m s -1 ;
to give h = 1.8 m;
gh31.8
A2. (a) above
(i) sea-level
use v 2 = 2=
;
m;
with v = 12 cos 60 = 6.0 m s -1 ;
or
v = 6.0 tomgive
s -1 ; h = 1.8 m;
(i)
M02/430/H(2)M+
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Markscheme
above sea-level = 31.8 m;
6.0
time to or
reach maximum height =
= 0.6s;
10
v = 6.0 m s -1;
1
1
6.0
=
vt
=
(6.0)
´ (0.6)
m;
height reached
time to reach maximum
height
= = 1.8
= 0.6s;
2
2
10
max height above sea =1 30 +11.8 = 32 m;
height reached = vt = (6.0) ´ (0.6) = 1.8 m;
2
2
max height above sea = 30 + 1.8 = 32 m;
(ii)
use v 2 = u 2 + 2 gh to find vertical speed with which stone hits sea with
u (ii)
= 6.0use
m s -v1;2 = u 2 + 2 gh to find vertical speed with which stone hits sea with
u = 6.0 m s -1;
-1
gives v = 25 m s ;
-1
v = 25=m
; 30 = 10 m s -1 ;
gives
12s cos
horizontal
speed
m s -1 ;
horizontal speed =212 cos 302 =110
2
{( 25) + (10) } ;
therefore speed = {( 25 ) + (10 ) }
therefore speed =
2
2
= 27 m s -1 ;
= 27 m s -1 ;
12t -5t 2 ; 2
or they might use -30 =-30
= 12t -5t ;
or they might use
or energy
argument;
or energy argument;
1
2
;
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M02/430/H(2)M+
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therefore speed =
{( 25)
2
+ (10 )
2
}
2
[1]
;
= 27 m s -1 ;
or they might use -30 = 12t -5t 2 ;
or energy argument;
Markscheme
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(b)
Velocity
Vertical
Horizontal
Time
correct vertical;
correct horizontal;
Do not penalise if the horizontal velocity is not shown to be greater than the
vertical velocity.
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