CONCEPT OF PROJECTILE.

Subject: 

Physics

Term:

FIRST TERM

Week:

WEEK 5

Class:

SS 2

Topic:

Previous lesson: 

The pupils have previous knowledge of

Equation of uniformly accelerated motion; (a) Revision on velocity-time graph (b) application and interpretation of equation of motion in simple problems.

that was taught as a topic in the previous lesson

Behavioural objectives:

At the end of the lesson, the learners will be able to

• Concept of projectiles
• Example of projectile motion
• Projectile of bodies at angle to the horizontal
• Horizontal projection

Instructional Materials:

• Wall charts
• Pictures
• Related Online Video
• Flash Cards

Methods of Teaching:

• Class Discussion
• Group Discussion
• Asking Questions
• Explanation
• Role Modelling
• Role Delegation

REFERENCES:

• new school Physics by MW Anyakoha
• New system PHYYSICS for senior secondary schools. Dr. Charles Chew.
• Comprehensive Certificate Physics by Olumuyiwa Awe
• Senior School Physics BY PN Okeke, SF Akande
• STAN Physics.

Content:

WEEK 5:

TOPIC; PROJECTILE

CONTENT

• Concept of projectiles
• Example of projectile motion
• Projectile of bodies at angle to the horizontal
• Horizontal projection

PERIOD ONE

CONCEPT OF PROJECTILE.

Projectile refers to the motion of a body which travels freely in space but under the influence of gravity and air resistance. When a ball in kicked into air, it will travel through space in a plane. The motion in a plane is a combination of upward and horizontal motion.

The path through which a projectile travel is called trajectory.

Example of projectile

In sport,

• Throwing of discus
• Throwing of javelin

In warfare

• Firing of catapult
• Shooting of arrows with bow
• Launching of missiles

Miscellaneous

• Throwing of stones

Projectile of bodies at angle to the horizontal

When a body is projected at an angle to the horizontal, the trajectory is a parabola.

The motion of this projectile can be splitted into two:

I. The horizontal motion

In the horizontal motion, the body moves with constant velocity. Therefore the horizontal acceleration is zero. This also implies that the initial and the final horizontal velocity are equal. a_x = 0V_x = U_x

U

U is the initial velocity with which the body was projected. Resolving U into it vertical and horizontal components, we have:

U

x

U

y

the horizontal Range R, is the horizontal distance covered by the projectile.

Since the acceleration along the horizontal is zero,

Horizontal Range R,

1. Vertical motion

The vertical motion is an example of a uniformly accelerated motion. The equations of uniform motion are still valid for it.

During the upward motion,

Vertical acceleration a_y = -g (where g is the acceleration due to gravity) Initial vertical velocity U_y = U sin

At the maximum height, the body is temporarily at rest. Therefore V_y = 0

Substituting these into V = U + at

V_y = U_y + at

t is the time to reach the maximum height.

The total time of flight is twice the time to reach the maximum height

Total time of flight T, — — — — (iii)

Using the equation,

Summarily

The horizontal range R can also be expressed as

For range to be maximum, 2must be equal to 90⁰_.

Therefore maximum range occur when

CLASSWORK:

1. a projectile is fired from the ground level with a velocity of 500m/s at 30⁰ to the horizontal. Determine;

• it horizontal range
• the greatest height attained.

SOLUTION:

R=?

R= 25000 x 0.866

R = 2165m

(i)

H_max = 6250m

1. A bullet is fired at an angle of 45⁰ to the horizontal with a velocity of 450m/s.

calculate (i) time to reach the maximum height (ii) the maximum height reached and the horizontal distance from the point of projection at this instant. SOLUTION t = 31.8 s

U

=

U

x

(Further examples should be solved as classwork)

EVALUATION

A particle is projected from the ground level with a velocity of 40m/s at an angle of . Calculate the

1. Time of flight
2. Range
3. Ime taken to reach the greatest height IV. Greatest height

PERIOD TWO

BODIES PROJECTED HORIZONTALLYAT A HEIGHT ABOVE THE GROUND

The motion of such projectile can also be splitted into two: the horizontal and the vertical motion.

1. Horizontal motion

In the horizontal motion, the body moves with constant velocity. Therefore, the horizontal acceleration is zero. This also implies that the initial and the final horizontal velocity are equal.

U_x = Vx = U a_x = 0.

If t is the time to reach the ground,then

1. Vertical motion

The vertical motion is an example of a uniformly accelerated motion. The equations of uniform motion are still valid for it.

During the upward motion,

Vertical acceleration a_y = -g (where g is the acceleration due to gravity)

The body was given an initial horizontal velocity. Since no vector has a perpendicular component, U_y = 0

Height H = S_y

Substituting these into S = Ut + ½ at².

CLASSWORK

1. A ball is projected horizontally from the top of building with a velocity of 10m/s. the height of the building is 45m. determine;

i. Time taken by the ball to reach the ground ii. Distance of the ball from the building after hitting the ground iii. The direction of the ball to the horizontal just before it hit the ground.

EVALUATION

1. For a particular value of U, at what to the horizontal should a ball be projected in order to have a maximum range? (a) 15⁰ (b) 30⁰ (c) 45⁰ (d) 60⁰
2. Which of these is not true about the horizontal motion of a projectile? (a) constant acceleration (b) uniform velocity (c) it is not affected by gravity (d) accelaeration is zero
3. A coin is pushed from the edge of a laboratory bench with a horizontal velocity of 15m/s of the height of the bench is from the floor is 1.5m. calculate the distance from the foot of the bench of the point of impact with the floor. (g = 10m/s²) (a)

0.75m (b) 2.25m (c) 8.22 m (d) 15.00m. (WASSCE 2011)

WEEKEND ASSIGNMENT

1. Using any of the equation of uniform motion, such that the maximum height attained by a ball projected at an angle to the horizontal with a velocity U is

READING ASSIGNMENT.

Students should answer the question on page 134 of New School Physics by MA Anyakoha