This document discusses projectile motion and provides equations to model the motion. It states that: 1) The motion of a projectile can be thought of as the combination of horizontal and vertical motions, which act independently. 2) The equation of motion for the horizontal direction is x = ut cosθ, where u is the initial velocity and θ is the launch angle. 3) The equation of motion for the vertical direction is y = utsinθ - 1/2gt^2. 4) Combining the horizontal and vertical equations yields the equation of the projectile path, which is a parabola.

1. NEHRU NAGAR
Prepared by:-
Siddhant-[XI-A] Saurabh-[X-C]

3. What is Projectile Motion?
 The motion of a projectile may be thought of as
the result of horizontal and vertical
components.
 Both the components act independently

4. Projectile Given Angular
Projection

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

11. Examples of Projectile Motion
• Launching a Cannon ball

12. Frame of reference: Equations of motion:
y
X Y
v0
Uniform m. Accel. m.
ACCL. ax = 0 ay = g = -9.81
g
h m/s2
VELC. vx = v0 vy = g t
x
0 DSPL. x = v0 t y = h + ½ g t2

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
0s 73.1 m/s, right 19.6 m/s, up
1s 73.1 m/s, right 9.8 m/s, up
2s 73.1 m/s, right 0 m/s
3s 73.1 m/s, right 9.8 m/s, down
4s 73.1 m/s, right 19.6 m/s, down
5s 73.1 m/s, right 29.4 m/s, down
6s 73.1 m/s, right 39.2 m/s, down
7s 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

18. Applications

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.

27. For your co-
operation