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Physics http://www.sigmaprc.in
Motion in a straight line
Freely falling body &
Vertically projected body
Physics http://www.sigmaprc.in
Motion in a straight line
Free falling body
Body projected from a tower
Vertically projected body
Free fall and vertical projection
Two freely falling bodies
Motion under gravity
Two vertically projected bodies
Physics http://www.sigmaprc.in
Motion in a straight line
Freely falling body
A body that is dropped and moves under the influence of only the gravitational forceis called a freely falling body
Important points regarding a freely falling body
Its initial velocity is zero ( u = 0 )
Acceleration is due to gravity ( 9.8 ms-2 and downwards )
The body gains velocity as is reaches the ground
It is a non-uniform motion with constant acceleration
Effect of air is assumed to be negligible
Quantities to analyze/determine
Time of descent
Final velocity of the body
Average velocity
Displacement in nth second
Physics http://www.sigmaprc.in
Motion in a straight line
Time of descent
It is the time taken by a freely falling body to reach the ground.
Consider a body dropped from a height H. Initial velocity ( u ) of the body is zero. Asthe body descends to ground, its velocity increases due to acceleration due togravity ( g ).
For a body moving with constant acceleration,displacement is given by
S ut at 21
2
Using u = 0 and a = - g ( as it is downwards )and S = -H ( also downwards ) we get
-H -g t 21
2
HTOD
g
25
Physics http://www.sigmaprc.in
Motion in a straight line
Final velocity
It is the velocity with which a freely falling body reaches the ground.
For a body moving with constant acceleration,velocity is given by
v u aS 2 2 2
Using u = 0 and a = - g ( as it isdownwards ) and S = -H ( alsodownwards ) we get
v gH 2
0v g H 2 2(- )(- )
6
Physics http://www.sigmaprc.in
Motion in a straight line
Average velocity
Average velocity is given by
Sv
t
avg
0
0
Hv
TOD
avg
Hv
H
g
avg
2
Hgv avg
2
Using the expression for TODwe get
Average velocity ( in case ofconstant acceleration only ) is alsogiven by
v avg2
u + v
0 ( 1) gHv
avg
2
2
+
Hgv avg
2
Alternate approach
Physics http://www.sigmaprc.in
Motion in a straight line
Displacements
Displacement in nth second of descent
S u a
n
1n
2
Using u = 0 and a = - g ( as it isdownwards )
S g
n
1n
2
Ratio of displacements in 1st , 2nd , 3rd seconds
gS g
1
11 1
2 2n = 1
gS g
2
12 3
2 2n = 2
gS g
3
13 5
2 2n = 3
: : : :S S S 1 2 3 1 3 5
Physics http://www.sigmaprc.in
Motion in a straight line
Displacements
Ratio of displacements in 1, 2, 3 seconds
S ut at 21
2
Using u = 0 and a = - g ( as it isdownwards ) we get
S gt 21
2
S g S g 21 11 (1)
2 2t = 1
S g S g 21 12 (2) 4
2 2t = 2
S g S g 21 13 (3) 9
2 2t = 3
: : : :g g g
S S S (1) (2) (3) 1 4 92 2 2
: : : :S S S (1) (2) (3) 1 4 9
( Motion in a straight line ( freely falling body )
Physics http://www.sigmaprc.in
Motion in a straight line
Vertically projected body
A body projected vertically up with an initial velocity executes1-D motion under the influence of gravitational force.
Important points regarding a vertically projected body
Its initial velocity is NOT zero
Acceleration is due to gravity ( 9.8 ms-2 and downwards )
The body loses velocity during ascent and gains velocityduring descent
It is a non-uniform motion with constant acceleration
Effect of air is assumed to be negligible
Quantities to analyze/determine
Time of ascent ( TOA )
Time of descent ( TOD )
Time of flight ( TOF )
The maximum height reached
Additional observations
Physics http://www.sigmaprc.in
Motion in a straight line
Time of ascent ( TOA )
Consider a body projected vertically up with an initial velocity u. During ascent itsvelocity decreases, becomes zero momentarily at the highest point, and thenincreases during descent. This is due to acceleration due to gravity ( g ).
Instantaneous velocity of a bodymoving with constantacceleration is given by
v u at
In this case a = - g ( as it isdownwards ) and at the highestpoint v = 0 therefore
uTOA
g 7
It is the time at which the velocity of the body becomes zeromomentarily
OrIt is the time taken by the body to reach the maximum height
( )u g t 0
ut
g
Physics http://www.sigmaprc.in
Motion in a straight line
Maximum height ( H )
Instantaneous velocity of a body movingwith constant acceleration is given by
It is the height at which the velocity of the body becomes zero momentarily
v u aS 2 2 2
Using v = 0 and a = - g ( as it isdownwards ) and S = H we get
uH
g
2
2
u g H 20 2(- )( )
8
Physics http://www.sigmaprc.in
Motion in a straight line
Time of descent ( TOD )
Considering the descent of the bodyfrom the highest point we get u = 0,a = - g and S = -H
It is the time taken by a vertically projected body to reach the ground , from theinstant when its velocity is momentarily zero.
S ut at 21
2
( )-H g t 210
2
H gt 21
2
Using the expression for maximum
height we getu
g
2
2
ugt
g
221
2 2
ut
g
22
2
ut
g
uTOD
g
Comparing this withequation (7) it is observedthat TOA = TOD.
Physics http://www.sigmaprc.in
Motion in a straight line
Time of flight ( TOF )
It is the time taken by a vertically projected body to reach the ground , from theinstant it is projected up.
uTOF
g
2
Since TOA is equal to TOD
TOF TOA 2×
uTOF
g 2 ×
( Motion in a straight line ( vertically projected body )
Physics http://www.sigmaprc.in
Motion in a straight line
Additional observations
Total displacement of the body is zero
Total distance covered by the body is 2H
Average velocity for the complete trip is zero
Average speed for the complete trip is u/2
Ascent and descent are symmetric
Distance covered by the body in 1st second of its ascent is equal to the distance
covered by it in the last second of its descent
Distance covered by the body in the last second of its ascent is equal to the
distance covered by it in the 1st second of its descent