equations projectile motion

8/13/2019 Equations Projectile Motion

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Motionin one dimension

Distance the length of the path betweeen two points.It is a scalarDisplacement the shortest distance between two points in a given direction. It is a vector

velocity =

displacement

time

speed =distance

time

Accelerationis the rate of change of velocity with time, so it is also a vector.

A change in SPEED, or

A change in DIRECTION, or

A change in speed AND direction

If an objects speed is constant but its velocity is changing, we say it is also accelerating

Average speed =Total distance

Total time

Average velocity =

Total displacement

Total time

Acceleration happens when there is:

Question: A man runs around a circular track of radius 100 m. It takes him 120 s to complete a

revolution of the track. If he runs at constantspeed, calculate:

1. his speed,

2. his instantaneous velocity at point A,

3. his instantaneous velocity at point B,

4. his average velocity between points A and B,

5. his average speed during a revolution.

6. his average velocity during a revolution.

N

W E

100 mDirection the man runs

b

A

b B

S

UNIT 1 PHYSICS ON THE GO

inal velocity initial velocity

average acceleration =time taken for change =

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Equations of motion

u = initial velocity (ms1) at t = 0 sv= final velocity (ms

1) at time t

s = displacement (m)

t = time (s)

a = acceleration (ms2)

v = u+ at

s =(u+ v)

2t

s = ut +12 at2

v2 = u2 + 2as

Question: A motorcycle, travelling east, starts from rest, moves in a straight line with a constantacceleration and covers a distance of 64 m in 4 s. Calculate

a) its acceleration

b) its final velocity

c) at whattime the motorcycle had covered half the total distance

d) what distance the motorcycle had covered in half the total time

conditions to use equations of motion

1. Acceleration should be constant

2. Motion should be along a straight line.

Derivarion of equations of motion

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Projectile motion - motion under gravity

1. Object thrown vertically

No horizontal component - consider vertical motion only.

In calculations, acceleration is taken-9.81ms-2if thrown upwardsand +9.81ms-2if downwards

If the object is thrown upwards, it continues to rise until v = 0

Question: A ball is thrown upwards with an initial velocity of 10 ms1.

1. Determine the maximum height reached above the throwers hand.

2. Determine the time it takes the ball to reach its maximum height.

Question: A cricketer hits a cricket ball from the ground so that it goes directly upwards. If the ball

takes, 10 s to return to the ground, determine its maximum height.

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-9.81 ms-2

+ 9.81 ms-2

v = 0

u = max

u

v= max


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2. Object projected horizontally

Initial velocity has no verticalcomponent, only a horizontal one

Acceleration is g= 9.81 ms-2vertically; no horizontal acceleration

The horizontal velocity is not altered by acceleration downwards. Neither is the accelerationdownwards changed by the horizontal motion.They are independent.

Final velocity

All objects launched horizontally or dropped from the same height will hit the ground at the same time.

Their vertical component of velocity is the same at all times.

Range (x)

Height (h)

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Question

1. A dartboard is 3.0 m away. A dart is thrown horizontally from height 1.90 m and hits the board at 1.60 m.

Calculate the

(a) time of flight

(b) initial speed of the dart

(c) speed of the dart when it hits the board.

2.A boy throws a stone horizontally off a cliff. It hits the sea 2 seconds later, at a distance of 40 m from

the foot of the cliff. Calculate the

(a) height of the cliff

(b) initial speed of the stone

(c) direction in which the stone is moving when it strikes the water.

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3. Launched at an angle to the horizontal

Initial velocity has both horizontal and vertical components

Acceleration is g= - 9.8 ms-2vertically; no horizontal acceleration

Object continues to move until it falls to the ground.

Question : A skater of mass 60 kg leaves the ice with a velocity of 10 ms1at an angle of

25 to the horizontal.

(a) Show that the vertical component of the skaters velocity is approximately 4 ms1.

(b) Calculate the time taken to reach the top of the jump.

(c) Calculate the maximum height reached.

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x

equations projectile motion

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