1. Presented by: Hasmuddin ansari
Projectile motion
Presented by:
Hasmuddin ansari

2. Some definitions:
Projectile: A body thrown into space which falls under action of gravity neglecting opposing air friction is called a projectile.
Projectile motion: The motion of a body thrown into space which falls under the action of gravity neglecting opposing air friction is called a projectile motion.

3. It is two dimensional motion.
For example: a stone thrown into space is a projectile and its motion is a projectile motion.
Trajectory: The path on which a projectile moves is called a trajectory. The trajectory of projectile is always a parabola. The direction of velocity is tangential to path.
Time of flight: The time taken by projectile to reach to ground from point of projection is called time of flight.
The time taken by projectile to reach to maximum height from ground is called time of ascent.

4. The time taken by projectile to reach to ground from maximum height is called time of descent.
Time of ascent and descent are equal.
Horizontal Range: The horizontal distance between point of projection and a point of strike on the ground by a projectile is called horizontal range.

5. To analyses projectile motion
Always find the components of velocity of projectile
Its velocity along horizontal is always constant.
Its acceleration along horizontal is zero.
i.e. ax = 0, ay = -g (up), ay = +g ( down)
Its time along horizontal, vertical and on path remains same at a point.
Always use the kinematical equation for horizontal and vertical motion.

6. Trajectory of projectile:
Consider a body thrown horizontally with speed u from the top of a tower of height h and it falls only under the action of gravity, neglecting opposing air friction. So,
Horizontal component of = ux = u cosθ = u coso0 = u
Vertical component of = uy = u sinθ = u sinoo = 0
Let the body reaches to any point P(x,y) in time t on its path where it covers horizontal distance x and vertical distance y respectively.
Now, considering horizontal motion:

7. sx = ux t + ½ ax t2 x = u t + ½. t2 t = x/u ………………
sx = ux t + ½ ax t2 x = u t + ½ .0. t2 t = x/u ………………. (i) And cosidering vertical motion: sy = uy t + ½ ay t2 y = 0.t + ½ g t2 y = ½ g (x/u)2 y = (½ g/u2) x2 = k x2 …………. (ii) Where k = (½ g/u2) = constant. This equation is a standard equation of parabola. Hence the trajectory of projectile is a parabola.

8. Alternatively,
Consider a body thrown horizontally with speed u at an angle θ from from ground and it falls only under the action of gravity, neglecting opposing air friction. So,
Horizontal component of = ux = u cosθ
Vertical component of = uy = u sinθ
Let the body reaches to any point P(x,y) in time t on its path where it covers horizontal distance x and vertical distance y respectively.
Now, considering horizontal motion:
sx = ux t + ½ ax t2
x = ucosθ t + ½ .0. t2
t = x/ucosθ ………………. (i)

9. And cosidering vertical motion: sy = uy t + ½ ay t2 y = usinθ
And cosidering vertical motion: sy = uy t + ½ ay t2 y = usinθ.t + ½ (-g) t2 y = usinθ.(x/ucosθ) - ½ g (x/ucosθ)2 y = x tanθ - ½ (g sec2θ/u2) x2 …………. (ii) This equation is a equation of parabola. Hence the trajectory of projectile is a parabola.

10. Derivation of time of flight, horizontal range and velocity of body at any time when thrown horizontally from top of a tower:
Consider a body thrown horizontally with speed u from the top of a tower of height h and it falls only under the action of gravity, neglecting opposing air friction. So,
Horizontal component of = ux = u cosθ = u coso0 = u
Vertical component of = uy = u sinθ = u sinoo = 0
Let the body reach to ground in time T.
Time of flight: Considering vertical motion,
sy = uy t + ½ ay t2
h= 0.T + ½ g T2

11. T =
This is the expression of time of flight.
Horizontal range (R): let the body strikes the ground at distance R from foot of tower in time T. So,
considering horizontal motion:
sx = ux t + ½ ax t2
R = u. t + ½ .0. t2
R = u .
This is the expression for horizontal range.
Velocity at any instant: let be the velocity of body in time t on its path.

12. Here , horizontal component of = vx = ux+axt =u+0.t= u
and vertical component of = vy = uy + ay t = 0 + gt = gt
So , magnitude of = v = =
and direction of = =
Derivation of time of flight, horizontal range and velocity of body at any time when thrown horizontally from ground at an angle θ .
Consider a body thrown horizontally with speed u at an angle θ from ground and it falls only under the action of gravity, neglecting opposing air friction.

13. So, Horizontal component of = ux = u cosθ
Vertical component of = uy = u sinθ
Let the body reach to ground in time T.
Time of flight: Considering vertical motion,
sy = uy t + ½ ay t2
0 = usinθ.T + ½ (-g) T2
T =
This is the expression for time of flight.
Time of ascent (ta) = ½ x time of flight =
Horizontal range (R): let the body strikes the ground at distance R from foot of tower in time T. So,
considering horizontal motion:

14. sx = ux t + ½ ax t2 R = ucosθ T + ½. T2 R = ucosθ
sx = ux t + ½ ax t2 R = ucosθ T + ½ .0. T2 R = ucosθ. = This is the expression for horizontal range. This shows that for a given initial velocity, the range of the projectile depends upon the angle of projection θ. The range R becomes maximum when sin2θ = 1 or, sin2θ = sin900 or, 2θ = 900 or, θ = 450

15. Hence , the projectile attains maximum range if it is thrown making an angle of 450 with the horizontal and the maximum horizontal range (Rmax) = u2/g .
Velocity at any instant: let be the velocity of body in time t on its path. Here ,
horizontal component of = vx = ux+axt = ucosθ
and vertical component of = vy = uy + ay t = usinθ - gt
So , magnitude of = v = =
and direction of =

16. Expression of maximum height and relation with Rmax:
Consider a body thrown horizontally with speed u at an angle θ from ground and it falls only under the action of gravity, neglecting opposing air friction. So,
Horizontal component of = ux = u cosθ
Vertical component of = uy = u sinθ
Let it reaches to maximum height (Hmax) and gain velocity .
Then , vertical component of = vy = 0
Now , considering vertical motion,
vy2 – uy2 = 2ay sy
or, 02 – (usinθ)2 = 2 (-g) Hmax

17. Hmax = u2/4g = ¼ (u2/g) Hmax = ¼ Rmax
Hmax = …………(i)
This is the expression for maximum height .
We have at angle of projection θ = 450 ,
The maximum horizontal range ( Rmax)= u2/g
So, at θ = 450 equation (i) becomes,
Hmax =
Hmax = u2/4g = ¼ (u2/g)
Hmax = ¼ Rmax
This is the relation between maximum height and maximum horizontal range.

18. Relation between two angle of projections for same horizontal range:
Let θ1 and θ2 be the two different angle of projections for a body projected with same speed u from ground which covers horizontal range R1 and R2 respectively.
So , R1 = u2 sin2θ1 /g and R2 = u2 sin2θ2/g
If R1 = R2 , we can write:
u2 sin2θ1 /g = u2 sin2θ2/g
or, sin2θ1 = sin
or, sin2θ1 = sin(180-2θ2 )
or, 2θ1 = θ2
or, θ1 + θ2 = 900
This is the relation between two different angle of projections for which horizontal range remains same.