MOTION UNDER GRAVITY IN TWO DIMENSION,DERIVATION

THIRD TERM E-LEARNING NOTE

SUBJECT: FURTHER MATHEMATICS CLASS: SS 2

SCHEME OF WORK

WEEK TWO

TOPIC:PROJECTILES: MOTION UNDER GRAVITY IN TWO DIMENSION,DERIVATION AND APPLICATION OF EQUATIONS INVOLVING GREATEST HEIGHT, TIME OF FLIGHT AND RANGE

Motion Under Gravity in Two Dimensions:

If a particle is projected with an initial velocity u at angle to the horizontal, the prativle will be resolved into vertical and horizontal components of the velocity.

V_Y

Horizontal components: Vx = ucos

Horizontal distance: Sx = utcos

Vertical components: Vy = usin

Vertical distance: sy = utsin – ½ gt²

Magnitude of the velocity, v = v_x ² + v_y²

The acceleration due to gravity acts against the motion of the body, hence it is negative.

Example:

A particle is projected with an initial velocity of 46m/s at an angle of 55⁰ to the horizontal. After 3 seconds, find: (i) the vertical component of the velocity (ii) horizontal component (iii) vertical distance traveled. (iv) Magnitude of the velocity.

SOLUTION:

= 55⁰ u=46m/s

(i) Vy = usin- gt

= 46 sin 55 – 10 x 3

= 37.68 – 30

= 7.68/s

(ii) Vx = ucos

= 46 cos 55

=26.38m/s

(iii) sy = utsin – (10×9)

= 138sin55 – 5×9

= 113.04 – 45

= 68.04m

(iv) = vx² +vy²

7.68² + 26.38² = 58.98 + 695.9

= 27.48m/s

EVALUATION: A particles is fired with an initial speed of 40m/s at an angle of 30⁰ to the horizontal. Determine the vertical and horizontal components of the velocity after 2.5 seconds.

GREATEST HEIGHT REACHED: when a projected particle reaches its greatest height, the vertical components become zero. Therefore;

RecallVy, = usin – gt

Squaring both sides, (Vy) ² = (usinn – gt)²

Vy^{2 =} u²sin² – 2gs_y

Since: vy =0, hence, 0 = u²sin² – 2gs_y

S_y =u²sin²

2g

Therefore the greatest height reaches is represented by H=u²sin²

2g

Time taken to reach the greatest height: The time taken to reach the maximum height I at the point when the vertical component is zero. Hence,

,Vy = usin–gt

0 = usin–gt

T=usin

g

Example:

A particle is projected with velocity 56m/s at an angle of 60⁰ from a point O on a horizontal plane. The particle moves freely under gravity and hits the plain again A. Calculate, correct to 3 significant figures: (a) the greatest height above OA attainedby the particle (b) the time taken by the particle to reach A from O.

Solution:

U = 56m/s = 60⁰

Greatest height reached, h = U²sin²O

2g

h = 56² x (sin 60)²

2 x 9.8

h = 2352 h = 120m.

19.6

(a) Time taken to reach A from O; t = usin

g

t = 56 sin 60

9.8

T = 4.9secs.

Evaluation: A project is fired with a velocity of 45m/s and at angle of elevation of 81⁰ to the horizontal. Find the time taken by the particle to reach its destination. (Take g = 10m/s²)

Time of flight: This is the time taken by a particle which is projected to return to its original point of projection. At this point the vertical distance becomes zero. Hemce,

T = 2usin

g

Range: This is the horizontal distance covered when the particle returns to its original point of projection. The range is equal to the product of the horizontal component and the time of flight.

Hence,

R =ucosx2usin

g

R =u² x 2sincos (but; 2sincos = sin 2)

g

R = u² x 2sin

g

Maximum range: A particle will cover a maximum range if it is projected at angle 45⁰ to the horizontal. That is; = 45⁰. Thus sin2 =1

Hence, R_max= U²

g

Example: The vertical and horizontal components of the initial velocity of a projectile are 36m/s and 64m/s. find (i) initial velocity of the projectile (ii) the inclination to the horizontal at which the projectile was fired. (iii) the greatest height reached; (iv) the time of flight; (v) the horizontal range of the projectile.

Solution:

V_y = 36m/s V_x= 64m/s

VV_x² +V_y²

U = 64^{2 +} 36²; U = 73.43m/s

Inclination to the horizontal; ( the angle of projection)

Vx = u cos

64 = 73.43 cos

= cos^-1 (64/73.43); = 29.4⁰

Greatest height reached; h = U² sin

2g

h = 73.43² x (sin 29.4)²

2 x10

h = 5391.96 x 0.2410

20

h = 64.97m

Time of flight: T = 2usin

g

T = 2x 73.43 x sin 29.4

10

T = 7.2 secs.

Horizontal range: R = u² sin2

g

R = 73.43² x sin (2×29.4)

10

R = 461.2m

EVALUATION:A particle is projected into the air with a speed of 50m/s at an inclination sin^-1(3/5). Find the: (greatest height reached by the particles; (ii) horizontal range; (iii) time of flight

Reading Assignment

New Further Maths Project 2 page 262 -270.

GENERAL EVALUATION

1) A particle is projected with an initial speed of 45m/s at an angle of 35 to the horizontal, find the time it takes for the particle to (i) reach the highest level (ii) return to its original level

2) A particle is projected horizontally with a velocity of 40m/s from the top of a tower 80.5m above the level ground find how far from the bottom of the tower the particle when it hits the ground

3) A particle is projected into the air with a speed of 20m/s at an inclination 30 to the horizontal , find the (i) greatest height reached (ii) horizontal range (iii) time of flight

4) Show that a particle which is projected with a given velocity reaches its maximum range at an elevation of sin^-1 (2^1/2 /2)

WEEKEND ASSIGNMENT

The vertical and horizontal components of the initial velocity of a projectile are 36m/s and 64m/s respectively find the

1) greatest height reached a) 32.4m b) 97.2m c) 64.8m d) 16.2m

2) time of flight a) 7.2s b) 3.6s c) 1.8s d) 14.4s

3) horizontal range a) 23.04m b) 46.08m c) 11.5m d) 92.16m

4) initial velocity of the projectile a) 73.4m/s b) 146.8m/s c) 36.7m/s d)18.4m/s

5) inclination to the horizontal a) 19 b) 21 c) 29 d) 49

THEORY

1) Find the initial speed which a projectile must be subjected to give a maximum horizontal range of 490m

2) Prove that the maximum range on a horizontal plane of a particle fired with velocity V at an angle x to the horizontal is V² / g