UNIT2D

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2.4 PROJECTILE MOTION

LESSON 4

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• To describe and use equations of projectile motion,

vx=vo cos ; vy=vo sin ax=0; ay=g=-9.81m s-2

• To derive equations for maximum height, H, horizontal range, R and time of flight, T.

Objectives:

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• Projectile motion is a combination of horizontal motion with constant velocity and vertical motion with constant acceleration

• Horizontal motion:

- velocity in the x-direction is constant ; vx = constant, ax = 0.• Vertical motion: - velocity in the y-direction has a constant acceleration of g (air resistance is

ignored) ; ay = g = -9.81 m s-2.

2.4 : Projectile Motion

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Parabola – the path is symmetrical

Where ; vo = initial velocityv = velocity at time tvy = velocity of y-component at time tvox = initial x-component velocityvoy = initial y-component velocitysx = horizontal rangeh = maximum height

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Equations of projectile motionEach vector quantity is resolved into 2 components - x component

- y component

Quantity Horizontal motion (x-component)

Vertical motion (y-component)

Displacement sx = voxt sy = voyt - ½ gt2

Velocity vx = vox (constant

velocity)

vy = voy - gt

Acceleration ax = 0 ax = g = 9.81 m s-2

(constant acceleration)

v0

v0x

v0y

x – component ; v0x = v cos

y – component ; v0y = v sin

ox

oy

v

v1tan

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Velocity at any point;

vx

vy

Equations of projectile motion

22v ; Magnitude yx vv

x

y

v

v1-tan ;Direction

v

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Maximum Range and HeightMaximum Range and Height

An object is launched at an initial velocity v0 at an angle with the horizontal.

H

Sx= R

x

y vy= 0

sy =0)

v0

Maximum Height , H

Vertically:

Initial velocity, v0y= v0 sin ,

At maximum height, vy = 0,

Using

vy2 = vo

2 + 2as

0 = (v0sin )2 + 2 (-g)H

2 20 sin

2

vH

g

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Vertically:

At the highest point, vy =0,

Using vy = voy – gt ;

0 = v0 sin - gt

The time taken to reach

the maximum height,

Time of flight, T

H

Sx= R

x

y vy= 0

sy =0)

v0

( sin )ovt

g

2 sin2 ov

T tg

As the path is symmetrical, time to go up = time to come down.

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Horizontally: ax= 0 ; vox = vo cos

sx = v0xt;

20 sin 2v

Rg

Range, R

The range is the horizontal displacement for the projectile.

R is maximum when 2 = 900

where sin 2 = 2 sin cos

90sin

2

max

2

g

vR

g

vR

o

oo

H

Sx= R

x

y vy= 0

sy =0)

v0

(2 sin )( cos ) o

o

vR v

g

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Maximum Range and Height• What are the conditions that give maximum

height and range in a projectile motion?

g

vh ii

2

sin 22 Maximum height can be achieved when i=90o!!!

2 sin 2i ivR

g

Maximum range can be achieved when 2i= 90o, i.e., i=45o!!!

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a) the maximum height ,Hb) the time of flight,c) the horizontal range, R,d) its velocity when it strikes the ground,

Example 1 – Projectile motion

A ball is launched with a velocity 20 m s-1 at an angle 60o with the horizontal at a point A. Calculate

a) H ?

c) Sx= R ?

x

y

Sy =0b) t ?

A

d) v ?

600

v0 = 20 ms-1

vh = 0

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a) H ?

c) Sx= R ?

x

y

Sy =0b) t ?

A

d) v ?

600

V0 = 20 ms-1

vh = 0

Solution of example 1 – Projectile motion;

A ball is launched with a velocity 20 m s-1 at an angle 60o with the horizontal at a point A.

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(a) Maximum height, H

Vertically: voy= 20 sin 600 ; vy = 0 (at maximum height)

Using vy2 = voy 2 – 2gs

0 = (20 sin 600)2 - 2(9.81)(H) H = 15.29 m

Solution of example 1 – Projectile motion

(b) Time of flight, T

Vertically:

voy= 20 sin 600 ; sy=0 ( Back on ground)

Using sy = v0yt - 1/2 gyt2

0 = v0 sin t - ½ gt2

Time of flight, T = 2v0 sin / g T = [2(20 sin 600 )]/9.81 = 3.53 s

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(c) Range, R

Horizontally :

vox= 20 cos 60o, T = 3.53 s, ax= 0

Using sx= voxt

R = (20 cos 60o)(3.53)

= 35.3 m

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(d) Velocity when the ball strikes the ground ;

Horizontally :

vx = vox = 20 cos 60o = 10 ms-1

Vertically :

v0y = 20 sin 60o ; T = 3.53s

Using vy = voy -- gt

vy= 20 sin 60o - (9.81)(3.53)

= - 17.31ms-1. (downward)

vx= 10 m s-1

vy = - 17.31m s-1 v


-12222 s m 2031.1710 yx vvv

o

x

y

v

v60

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31.17tantan 11

Magnitude:

from the diagram, v = -20 m s-1 (downward)

Direction :

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At an altitude of 500 m, a bomb is released from an airplane flying horizontally with uniform velocity, 100 m s-1 to hit a target which is located 800 m away. Find the time the bomb reaches the ground.Is the mission successful?

500 m

800 m


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x

y

vo = 100 ms-1

sy = -500 m

t ?

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Vertically:voy = 100 sin 0o = 0 ; g = 9.81m s-2 ; sy = -500 m.

sy = voyt - 1/2 gt2

-500 = 0 - 1/2 gt2

t = 10.1 s

Horizontally:vox = 100 cos00 = 100 m s-1 ; ax = 0 ; t = 10.1 s

sx = voxt - 1/2 axt2

= (100) (10.1) + 0 = 1010 m Since the target is located 800 m, hence the bomb misses the

target and so the mission is not successful.


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A firefighter, a distance d from a burning building, directs a stream of water from a fire hose at angle i above the horizontal as in the figure. If the initial speed of the stream is vi , at what height h does the water strike the building?


Solution :

The horizontal component of displacement,

cosx ix i is d v t v t the time required to reach the building a distance d away is ;

cosi i

dt

v

At this time, the altitude of the water ; 21

2y iyv h v t gt 2

sincos 2 cosi i

i i i i

d g dv

v v

the water strikes the building at a height h above ground level of ;2

2 2tan

2 cosy ii i

gds h d

v

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Example 4 – Projectile motionA ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m s-1, 40o above horizontal. How far above or below its original level will the ball strike the opposite wall ?

)40o

20 m s-1

50 m

Solution :0 -120cos 40 15.3 m soxv 0 -120sin 40 12.9 m soyv

Horizontally ; x x oxs v t v t

503.27 s

15.3t

Vertically ;

212y oys v t gt

212(12.9)(3.27) (9.81)(3.27)

10.27 m

Therefore, the ball hit at 10.27 m below the original level.

time taken to strike the opposite wall :

original level

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Example 5 – Projectile motionA projectile is fired horizontally with a speed of 30 m s-1 from the top of a cliff 80 m high. a) How long will it take to strike the level ground at the base of the cliff ? b) How far from the foot of the cliff will it strike ?

c) With what velocity will it strike ?

80 m

30 m s-1 Solution :

80 m ;ys -130 m s ;o oxv v -10 m soyv

a) Vertically ; 212y oys v t gt

21280 (0) (9.81)t t

4.04 st b) Horizontally ;

x oxs v t(30)(4.04) 121.2 mxs c) Final velocity , v ;

-130 m s ;x oxv v -10 (9.81)(4.04) 39.6 =40 m sy oyv v gt

2 2 2 2 -1Magnitude, 30 ( 40) 50 m sx yv v v

-1 040Direction , = tan 53

30

-1 0Hence, v = 50 m s ; 53 below +ve x-axis

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A boy playing football kicks the ball at an angle of 37o with respect to the horizontal, at an initial speed of 15 m s-1. Assuming that the ball moves in a vertical plane. Calculate ;

a) the time t1 after which the ball will reach the highest point of its trajectory and the height of this point above the ground

b) the time t2 taken by the ball to fall on the ground and its speed at that moment

c) the longest possible range of the ball


Answer :

a) t1 = 0.9 s ; ymax = 4.1 m

b) t2 = 1.8 s ; v = 15 m s-1

c) xmax = 22.9 m

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Conclusion

2 20 sin

2

vH

g

20 sin 2v

Rg

• Equations of projectile motion, vx= vo cos ; vy= vo sin ax= 0; ay= g = -9.81m s-2

• Equations for maximum height, H, horizontal range, R and time of flight, T (for a parabolic curve)

g

vT o sin2

H

R

y

x

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