This document discusses rectilinear motion under gravity. It defines motion under gravity as motion of a particle projected vertically in air under the influence of gravitational force. It presents the equations of rectilinear motion and modifies them for motion under gravity. It notes that time of flight is the total time an object remains in air, which is the sum of time for upward and downward journeys. It provides examples calculating the time and velocity at which two objects crossing paths under gravity conditions.

1. Unit-5
Rectilinear Motion of Particle
Sub- Engg. Mechanics
By: Mr. Ghumare S. M.
Motion Under Gravity
 When the particle projected vertically in the air, then its motion
is under the gravitational force, this motion is called as Motion
under Gravity.
Here,
2
9.81 / seca g m 

2. Motion Under Gravity
• Equations of Rectilinear Motion
 
2
2 2
1
1
(2)
2
2 (3)
v
s
v
u a t
u t a t
u a s
         
       
       
Equations for Motion Under Gravity modified from above
equations
 
2
2 2
, ,
1
1
(2)
2
2 (3)
Here
v
s = h
v
a g s h
u g t
u t g t
u g h
 
         
     
       

3. Motion under Gravity
• Important Points
 When objects thrown vertically upward in the air, it attains the
maximum height at certain point where final velocity becomes
zero (V=0).
 When objects released from the top of tower or building vertically
downward, then its initial velocity at that point becomes zero.
 The time of travel required for upward and downward journey
for the object must be same.
Hence, time of flight = Total time for which object remained in air
= Time of upward plus downward journey

4. Motion under Gravity
• Important Points
 Assume acceleration due to gravity ‘g’ as positive for downward
journey and Negative for upward journey
 The velocity of the particle at height must be same in magnitude
in upward and downward motion/ direction.

5. Numerical Example-1
Ex. From, top of the tower 100m high , a stone is dropped down. At
the same time, another stone is thrown upward from the foot of the
tower with a velocity 30 m/sec. When and Where the two stones will
cross each other? Find the velocity of two stones at the time of
crossing.
Given, Ht. of tower 100m
Stone 1 dropped from top, Hence initial velocity = 0
Stone 2 thrown upward with initial velocity = 30 m / s
Let, ‘t’ be the time required for the stones during which both stones
will covers some distance, let us say and .
But, here
1u
2u
1h 2h
1 2 100 (1)h h m        

6. Example 1 Continue….
• Distance travelled by 1st Stone in downward
direction is as follows,
2
1 1
2
1 1
2
1
2
1
1
2
, , 0
1
2
1
, 9.81
2
4.905 (1)
s
is
then X X
u t g t
s h g ve u
h u t g t
h t
h t
 
  
 

     

7. Example 1 Continue….
• Distance travelled by 2st Stone in upward
direction is as follows,
2
2 2
2
2 2
2
2
2
2
1
2
, , 30 /
1
2
1
, 30 9.81
2
30 4.905 (2)
s
is
then X X
u t g t
s h g ve u m s
h u t g t
h t t
h t t
 
  
 
 
       

8. Example 1 Continue….
Total height of the tower is 100 m
In time t both the stones have traveled distance h1 and h2,
1 2
1 2
2 2
1 2
, .(1) (2)
100
100 4.905 30 4.905
100 30 , 3.33sec
.(1) (2),
54.5 45.5
9.81 3.33 , 32.7 / sec
st
1 1
1 1
To find velocity of crossing,
Use v =u +gt for 1 Stone
v = 0 + X v
Put h and h from Eq and
m h h
t t t
t then t s
putting in Eq and
h m and h m
m
 
  
 
 

2 2
2 2
( )
30 9.81 3.33 , 2.7 / sec( )
st
Use v =u -gt for 2 Stone
v = X v m

   

9. Numerical Example-2
Ball ‘A’ is released from the rest at a height 12 m rom top, At the
same time, another ball is thrown upward 1.5m from the ground. If
the balls cross each other at a height of 6m, Determine the speed at
which ball B was thrown upward.
Given, Ht. of tower 12m
Ball 1 dropped from top, at rest, initial velocity = 0
Stone 2 thrown upward 1.5m from ground, Find
Let, ‘t’ be the time required for the Balls during which both stones
will covers some distance, let us say and .
But, here
1u
2u
1h 2h
1 2 12 1.5 10.5 (1)h h m          

10. Example 2 Continue….
• Downward distance travelled by Ball A is 6m,
1 1
2
6 , , 0
1
6 0 9.81
2
1.106
iss h m g ve u
t
t Secs
   
 

Downward distance travelled by Ball B =4.5M
2 2
2
2
2
4.5 , , ?
1.106
1
4.5 1.106 9.81 (1.106)
2
9.49 / sec
iss h m g ve u
Put t Secs
X u X
Solving u m
   

 


11. Thank You