Equations of constant accelerations

7/29/2019 Equations of constant accelerations

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Equations of ConstantAcceleration

Applications of Newtons Laws of

Motion


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Objectives

Understand projectile motion and howgravity influences a projectile.

Understand the effects of projectile speed,relative height and projection angle onprojectile motion.

Learn to compute maximum height, flighttime and range using equations of uniformacceleration.


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Projectile

A projectile is a body or object that is in the

air.

Only under the influence of gravity and air

resistance.


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Gravity

Gravity influences a projectile by pulling

the object toward the earth.

Acceleration due to gravity = - 9.8 m.s-1

Gravity causes a projectile to follow a

parabolic path.


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Why are projectiles important?

Success of many sporting events involvesprojectiles

Objects acting as projectiles Basketball, discus, others?

People acting as projectiles

Variables of interest Flight distance (range)

Flight time

Maximum height


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Equations of Constant

Acceleration

d = vit + .a.t

vf= vi + a.t

vf = vi + 2.a.d

Where

d = displacement

vi. = initial velocity

vf= final velocity

a = acceleration


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Equations of Constant

Acceleration

The equations are most commonly applied

when a rigid body is in flight and air

resistance is ignored and therefore the onlyforce acting is gravity.


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Applications of Equations

If a trampolinist is in the air for two

seconds, how high does he/she jump?

Questions: Which equation might you use?


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What is the velocity of the athlete at the

very top of the activity?

Answer: vi = 0

What is the value for gravity?

Answer: g = -9.8 m.s.-2


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Therefore:

d = vi

.t + .a.t

d = (0.0)(1) + (-9.8)(1)

d = -4.905 m


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Application II

A figure skater is attempting a jump in which she

performs 3 revolutions while in the air. She leaves

the ice with a velocity of 7 m/s at a projectionangle of 30. How long will she be in the air?

Is she spins 3 revolutions per second, will she be

able to complete all 3 revolutions before landing?


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Applications of Equations II

A ball is kicked from the ground level at an

angle of 60 degrees to the horizontal at a

velocity of 20 m.s 1.

Ignoring air resistance, how far does it

travel?


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Resolve velocity vector into horizontal and

vertical components

Therefore, velocity in x direction is 10 m.s-1

Velocity in the y direction is 17.3 m.s-1


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Time of flight is determined by the velocity

in the up direction (y direction)

Which equation of motion could we use to

determine the time of flight?

Answer: vf= vi + a.t


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vi = 17.3 m.s-1

vf

= 0.0 m.s-1

a = -9.8 m.s-2

tup =1.76 s

tflight = 3.52 s


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Application of Equations II

What is the distance traveled horizontally?

What equation could we use?

Answer: d = vi.t + .a.t

There is no force acting

horizontally,therefore a = 0.0

d = 35.2 m


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Conclusion

The distance traveled depends on the time

in flight which depends on the vertical

velocity and horizontal velocity.

For a maximum distance to be obtained, a

high vertical and high horizontal velocity is

required


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The best compromise between the two

when resolving the velocity is at an angle of

45 degrees.


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Application of Equations III

Throwing

An athlete throwing from a projectile does

not release from ground height.

Therefore, there is a more complex

relationship between optimum angle of

release and height of release.


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The relationship is described by the

following equation.

R = (V/g).cos().


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As a general rule, the higher the release the

further the projectile goes.

As height of release increases, the angle of

release decreases.

For a shot putter releasing at a height of

2.44 m, the optimum angle is 40 degrees.


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Influences on Projectiles

Projection speed

Projection angle

Relative height

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Equations of constant accelerations