Projectile SS2 Further Mathematics Lesson Note

MEANING OF PROJECTILE 

A projectile motion follows a curved or parabolic path. It is due to two independent motions at a right angle to each other. These motions are:

1. horizontal constant velocity 
2. vertical  free fall due  to gravity 

Examples of projectile motion are the motion of;

1. thrown rubber ball are-bouncing from a wall
2. An athlete doing the high jump
3. A stone released from a catapult
4. A bullet fired from a gum
5. A cricket ball is thrown against a vertical wall.

Uy = U sin θ (vertical component) ——————- 1

Ux = U cos θ (horizontal component) ——————- 2

TERMS ASSOCIATED WITH PROJECTILE

1. Time of flight (T) – The time of flight of a projectile is the time required for it to return to the same level from which it is projected.

t= time to reach the greatest height 

V = u + at (but, v =o, a = -g)

θ= u sin – gt

t = U sin θ ———-   3

         g

T = 2t = 2U sin θ ——–  4

                    g 

2. The maximum height (H) – is defined as the highest vertical distance measured from the horizontal projection plane.

For maximum height H, 

V² = U² sin²θ – 2g H

At maximum height H, V=0

2gH = U² ———-   5

           2g 

3. The range (R) – is the horizontal distance from the point of projection of a particle to the point where the particle hits the projection plane again.

Horizontally, considering the range covered:

Using S= ut + ½at² (where a=0 for the horizontal motion)

OR

S = R = U cosθ x t (distance = velocity x time; their time is the time of flight)

R = U cos θ (2 U sin θ)

                           g

R = 2U² sin θ cos θ

                   g

From Trigonometry function

2 sin θ cos θ = sin 2θ

R= U² sin 2θ

           g

For maximum range θ = 45°

Sin2θ = sin 2 (45) = sin 90° = 1

R= U²

      g

Rmax = U²

               g 

 USES OF PROJECTILES

1. To launch missiles in modern warfare
2. To give athletes maximum takeoff speed at meets
3. In artillery warfare, to strike a specified target, the bomb must be released when the target appears at the angle of depression φ given by:

Tan φ =1/u  √gh/2  

EXAMPLES

1. A bomber on a military mission is flying horizontally at a height of zoom above the ground at 60 km in-1. Lt drops a bomb on a target on the ground. Determine the acute angle between the vertical and the line joining the bomber and the tangent at the instant. The bomb is released

Horizontal velocity of bomber = 60km/min= 103 ms-1

Bomb falls with a vertical acceleration of g = 10m/s

At the release of the bomb, it moves with a horizontal velocity equal to that of the aircraft i.e. 1000m/s

Considering the vertical motion of the bomb we have

h =ut+1/2 gt2 (u=o)

h =1/2gt2

Where: t is the time the bomb takes to reach the ground: 300=1/2gt2

t2= 600

t=10√6 sec

Considering the horizontal motion we have that the horizontal distance moved by the bomb in time t is given by

s =horizontal velocity x time

s = 1000 x10√6

s = 2.449×104 m

But tanθ =  s   =       2.449 x 104

                3,000            3,000

                θ =83.020 

2. A stone is shot out from a catapult with an initial velocity of 30m at an elevation of 600. Find:
3. the time of flight
4. the maximum height attained
5.  the range

1. The time of flight

T = 2U sin θ

           g 

T= 2 x 30 in 600

             10

T= 5.2s

2. The maximum height,

H=U² sin2 θ

         2g

H = 30² sin2 (60)

              20

H = 33.75 m

3. The range,

R = U²sin 2θ

           g

R = 30² sin 2 (60)

            10

R = 90 sin 120

R = 77.9 m