Sanjivani College of Engineering Kopargaon

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