Freely falling body Vertically projected bodysigmaprc.in/studymaterial/freefall-vertical.pdfPhysics...

Physics http://www.sigmaprc.in

Motion in a straight line

Freely falling body &

Vertically projected body



Free falling body

Body projected from a tower


Free fall and vertical projection

Two freely falling bodies

Motion under gravity

Two vertically projected bodies



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



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



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



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



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



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 )

http://www.sigmaprc.in/




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



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



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



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.



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 )

http://www.sigmaprc.in/



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

Freely falling body Vertically projected bodysigmaprc.in/studymaterial/freefall-vertical.pdfPhysics...

Documents

Transcript of Freely falling body Vertically projected bodysigmaprc.in/studymaterial/freefall-vertical.pdfPhysics...