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NEHRU NAGAR
Prepared by:-
Siddhant-[XI-A] Saurabh-[X-C]
 The motion of a projectile may be thought of as
the result of horizontal and vertical
components.
 Both the components act independently
What is Projectile Motion?
Projectile Given Angular
Projection
Equation of path of projectile
Suppose at any time t, the object is at point P (x, y).
For motion along horizontal direction, the acceleration ax is zero.
The position of the object at any time t is given by,
Here, x0 = 0, ux = u cos θ, ax = 0
[ Velocity of an object in the horizontal direction is constant]
Putting these values in equation (i),
⇒ x = ut cos θ
For motion along vertical direction, the acceleration ay is −g.
The position of the object at any time t along the vertical direction is
given by,
Here,
∴
Putting the value of t from equation (ii),
⇒
This is an equation of a parabola. Hence, the path of
the projectile is a parabola
Time of flight
Total time for which the object is in flight
It is denoted by T.
Total time of flight = Time of ascent + Time of descent
∴ T = t + t = 2t [Time of ascent = Time of descent = t]
At the highest point H, the vertical component of velocity
becomes zero. For vertical motion of the object (from 0 to H),
⇒
∴
Maximum height
•Maximum height ‘h’ reached by the projectile
For vertical upward motion from 0 to H,
Using the relation
we obtain
⇒
That is,
Horizontal Range
•Horizontal distance covered by the object between its point of projection and the point of hitting the
ground. It is denoted by R.
‘R’ is the distance travelled during time of flight T.
⇒
⇒
For the maximum horizontal range,
sin 2θ = 1 = sin 90°
⇒ 2θ = 90°
⇒ θ = 45°
∴ Maximum horizontal range (Rm
) is
Examples of Projectile Motion
• Launching a Cannon ball
x
y
0
Frame of reference:
h
v0
Equations of motion:
X
Uniform m.
Y
Accel. m.
ACCL. ax = 0 ay = g = -9.81
m/s2
VELC. vx = v0 vy = g t
DSPL. x = v0 t y = h + ½ g t2
g
Important points :
In case of projectile motion , the horizontal
component of a velocity (u cos θ) , acceleration
(g) and mechanical energy remains constant
while, speed ,velocity , vertical component of
velocity (v sin θ) ,momentum ,kinetic energy
and potential energy all change.
Velocity and K.E are maximum at the point of
projection while minimum (but not zero) at the
highest point.
Vector diagrams for
projectile motion
TIME HORIZONTAL VELOCITY VERTICAL VELOCITY
0 s 73.1 m/s, right 19.6 m/s, up
1 s 73.1 m/s, right 9.8 m/s, up
2 s 73.1 m/s, right 0 m/s
3 s 73.1 m/s, right 9.8 m/s, down
4 s 73.1 m/s, right 19.6 m/s, down
5 s 73.1 m/s, right 29.4 m/s, down
6 s 73.1 m/s, right 39.2 m/s, down
7 s 73.1 m/s, right 49.0 m/s, down
 What two factors would affect projectile motion?
◦ Angle
◦ Initial velocity
Initial Velocity
Angle
Evaluating various info
Determination of the Time of Flight
Determination of Horizontal Displacement
x = vix • t
Determination of the Peak Height
y = viy • t + 0.5 • g • t2
Applications
The use of projectile
motion in sports
 Like:-
Cricketers know that from which angle
they have to hit to get the maximum range.
For your co-
operation

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Projectile motion

  • 2.
  • 3.  The motion of a projectile may be thought of as the result of horizontal and vertical components.  Both the components act independently What is Projectile Motion?
  • 5. Equation of path of projectile Suppose at any time t, the object is at point P (x, y). For motion along horizontal direction, the acceleration ax is zero. The position of the object at any time t is given by, Here, x0 = 0, ux = u cos θ, ax = 0 [ Velocity of an object in the horizontal direction is constant] Putting these values in equation (i), ⇒ x = ut cos θ For motion along vertical direction, the acceleration ay is −g. The position of the object at any time t along the vertical direction is given by, Here,
  • 6. ∴ Putting the value of t from equation (ii), ⇒ This is an equation of a parabola. Hence, the path of the projectile is a parabola
  • 7. Time of flight Total time for which the object is in flight It is denoted by T. Total time of flight = Time of ascent + Time of descent ∴ T = t + t = 2t [Time of ascent = Time of descent = t] At the highest point H, the vertical component of velocity becomes zero. For vertical motion of the object (from 0 to H), ⇒ ∴
  • 8. Maximum height •Maximum height ‘h’ reached by the projectile For vertical upward motion from 0 to H, Using the relation we obtain ⇒ That is,
  • 9. Horizontal Range •Horizontal distance covered by the object between its point of projection and the point of hitting the ground. It is denoted by R. ‘R’ is the distance travelled during time of flight T. ⇒ ⇒ For the maximum horizontal range, sin 2θ = 1 = sin 90° ⇒ 2θ = 90° ⇒ θ = 45° ∴ Maximum horizontal range (Rm ) is
  • 10.
  • 11. Examples of Projectile Motion • Launching a Cannon ball
  • 12. x y 0 Frame of reference: h v0 Equations of motion: X Uniform m. Y Accel. m. ACCL. ax = 0 ay = g = -9.81 m/s2 VELC. vx = v0 vy = g t DSPL. x = v0 t y = h + ½ g t2 g
  • 13. Important points : In case of projectile motion , the horizontal component of a velocity (u cos θ) , acceleration (g) and mechanical energy remains constant while, speed ,velocity , vertical component of velocity (v sin θ) ,momentum ,kinetic energy and potential energy all change. Velocity and K.E are maximum at the point of projection while minimum (but not zero) at the highest point.
  • 14. Vector diagrams for projectile motion TIME HORIZONTAL VELOCITY VERTICAL VELOCITY 0 s 73.1 m/s, right 19.6 m/s, up 1 s 73.1 m/s, right 9.8 m/s, up 2 s 73.1 m/s, right 0 m/s 3 s 73.1 m/s, right 9.8 m/s, down 4 s 73.1 m/s, right 19.6 m/s, down 5 s 73.1 m/s, right 29.4 m/s, down 6 s 73.1 m/s, right 39.2 m/s, down 7 s 73.1 m/s, right 49.0 m/s, down
  • 15.  What two factors would affect projectile motion? ◦ Angle ◦ Initial velocity Initial Velocity Angle
  • 16. Evaluating various info Determination of the Time of Flight Determination of Horizontal Displacement x = vix • t Determination of the Peak Height y = viy • t + 0.5 • g • t2
  • 17.
  • 19. The use of projectile motion in sports
  • 20.  Like:- Cricketers know that from which angle they have to hit to get the maximum range.
  • 21.
  • 22.
  • 23.
  • 24.
  • 25.
  • 26.