lesson thread i full projectile motion

Intellectual Property of Mr. Gary 1

Lesson ThreadI

Full Projectile Motion

12/1/2013


Projectile Motion

Lesson I

Determine the Maximum Height

Part a

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A batter on the ladies softball team hits a fly ball with an initial velocity of 35 m/s at an angle of 27° above the horizontal. Calculate (a) the maximum height the ball reaches above the ground, (b) the x, y – components of the velocity of the ball at the maximum height, (c) the time it takes the ball to reach the maximum height, (d) the maximum range the ball traveled from the batter.

Question: Part a

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0v

0 0 0 cosxv v

0 0 0sinyv v

27

1 1 1 cosxv v 1 1 1sinyv v

27

2 2 2cosxv v

2 2 2sinyv v

0v

y��

Sketch: Part a

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212

2 20

sin sin

sin sin

( sin ) ( sin ) 2

gv v t

y v t g t

v v g y

��

��

��

0

212

2 2

cos cos

cos

( cos ) ( cos ) 2

v v a t

x v t a t

v v a x

Component Equations: Part a

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max max max

max

ma

20

x

20

Eq. 2.11c

Find the maximum height

( sin ) ( sin ) 2

The maximum vertical height has an angle 0

( sin(0 )

f f

f

f

v v g y

RPC

v

��

max

max

m

0 2 20

ax

0

20 0

2 20 00 0

) ( sin ) 2

0 ( sin ) 2

Perform all necesary algebraic steps to solve for y .

( sin ) 0 [ ( si )

n

v g y

v g y

v v

��

��

2

0

0

0( sin )v 0

20

max

ma0

20

x

0

] 2

( sin ) 0 2

2( sin )

2

g y

v g y

v

g

��

��

��

1g�� 1

max

2

y��

1g��

20 0

1

2

max

max

0 0

( sin )

2

( sin )

2

vy

g

vy

g

��

��

��

��20

(35 sin)(27 ))

2(

m s

209.8 )

12.88mm s

Solution: Part a

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Projectile Motion

Lesson IIDetermine the x, y – components at the

maximum height.Part b

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Question: Part b

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0v

0 0 0 cosxv v

0 0 0sinyv v

27


27

2 2 2cosxv v

2 2 2sinyv v

0v

y��

Sketch: Part b

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212

2 20

sin sin

sin sin

( sin ) ( sin ) 2

gv v t

y v t g t

v v g y

��

��

��

0

212

2 2

cos cos

cos

( cos ) ( cos ) 2

v v a t

x v t a t

v v a x

Component Equations: Part b

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1 1 1

1

Solving for the y-component

sin

(35 ) s 0in( )

y

y

v v

RPC

v m s

0

1 0y m sv

1 1 1

1

Solving for the x-component

cos

(35 ) c 0os( )

x

x

v v

RPC

v m s

1

1 35x mv s

Solution: Part b

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Projectile Motion

Lesson IIICalculate the time taken to reach the

maximum heightPart c

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Question: Part c

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0v

0 0 0 cosxv v

0 0 0sinyv v

27


27

2 2 2cosxv v

2 2 2sinyv v

0v

y��

Sketch: Part c

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212

2 20

sin sin

sin sin

( sin ) ( sin ) 2

gv v t

y v t g t

v v g y

��

��

��

0

212

2 2

cos cos

cos

( cos ) ( cos ) 2

v v a t

x v t a t

v v a x

Component Equations: Part c

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Solution: Part c

1 01 0

1

Choose Eq. 2.11a: y-component

sin sin

sin 0( )

y yy y

y

gv v t

RPC

v

��

00 0

0 00 0

sin

( sin ) 0 [ ( sin )

y y

y yy y

gv t

v v

��

0

0 0( sin )y yv 0

0 0

0 0

]

( sin )

( sin )

y y

y y

g

g

gt

t

v

tv

��

��

��

1(35 )sin(27 )

(

m s

219.8 )

1.62

m s

t s

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Projectile Motion

Lesson IVCalculate the Range the particle travels

horizontallyPart d

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Question: Part d

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0v

0 0 0 cosxv v

0 0 0sinyv v

27


27

2 2 2cosxv v

2 2 2sinyv v

0v

y��

Sketch: Part d

x = Range

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212

2 20

sin sin

sin sin

( sin ) ( sin ) 2

gv v t

y v t g t

v v g y

��

��

��

0

212

2 2

cos cos

cos

( cos ) ( cos ) 2

v v a t

x v t a t

v v a x

Component Equations: Part d

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212

210

0

20

0

Choose Eq. 2.11b

cos

There is no acceleration in x-direction

cos

x v t a t

x v t a t

0 0

0

0

0

cos 0

cos

We can now double the time calculated in Part b and substitute

that value into .

(35 )cos(27 ) (1.62s)

(35

t

2

x v t

RPC

x v t

x m s

x m s

1)cos(27 ) (3.24 s

1

101. 4

)

0x m

Solution: Part d

12/1/2013

lesson thread i full projectile motion

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