**The Physics of Archery**

Preview

History of Archery

20,000 B.C

Egyptians

Yew Longbow

Compound Bow

1/2 MaV^2=1/2Fx

**Collision**

M1V1+M2V2=(M1+M2)Vf

Inelastic Collision

M1V1+M2V2=M1Vf1+M2Vf2

Elastic Collision

Potential to Kinetic Energy

Time Aloft

Taloft=| 2Vy/ag |

**Work**

**W=FD**

The potential energy of an arrow is 2 joules. What is the velocity of the arrow if the mass is 1.5 g?

1/2MaV^2=1/2Fx

M1V1+M2V2=(M1+M2)Vf

Inelastic Collision

An arrow, weighing 1.5g, hits an apple, weighing 170 g, at 83 m/s and continues with the apple what is the final velocity?

M1V1+M2V2=M1Vf1+M2Vf2

Elastic Collision

An arrow, weighing 2.2 g, hits a stationary block, weighing 10 kg, at 80 m/s. If the blocks final velocity is 0.024 m/s, what is the final velocity of the arrow

Taloft=I 2Vy/ag I

W=Fd

An arrow is drawn back in a recurve bow a distance of 70 cm. If the work done is 77 Joules what is the force exerted on the arrow?

An arrow is shot at a 20 degree angle with a launch velocity of 78 m/s what is the time aloft?

Potential to Kinetic Energy

Inelastic and Elastic Collisions

Time Aloft

Work

**What Is Archery?**

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Inelastic Collision

An arrow, weighing 1.5g, hits an apple, weighing 170 g, at 83 m/s and continues with the apple what is the final velocity?

M1V1+M2V2=(M1+M2)Vf

1.5g = 0.0015 kg 170g = 0.17

M1= 0.0015 kg

V1= 83 m/s

M2=0.17 kg

V2= 0 m/s

Vf= ?

(0.0015)(83)+(0.17)(0)=(0.0015+0.17)Vf

0.1245=(0.1715)Vf

Vf= 0.726 m/s

1/2 MaV^2=1/2Fx

The potential energy of an arrow is 2 joules. What is the velocity of the arrow if the mass is 1.5 g?

1.5g = 0.0015 kg

Ma= 0.0015 kg

V= ?

Pe= 2 joules

Ke=Pe

1/2 (0.0015)(V^2)=2

(7.5*10^-4)(V^2)=2

V^2=2666.666

V= 51.639 m/s

An arrow, weighing 2.2 g, hits a stationary block, weighing 10 kg, at 80 m/s. If the blocks final velocity is 1 m/s, what is the final velocity of the arrow

M1V1+M2V2=M1Vf1+M2Vf2

Elastic Collision

2.2g = 0.0022 kg

M1=0.0022 kg

V1= 80 m/s

M2= 10 kg

V2= 0 m/s

Vf1= ?

Vf2= 0.024 m/s

(0.0022)(80)+(10)(0)=(0.0022)(Vf1)+(10)(0.024)

0.176=(0.0022)(Vf1)+0.24

-0.064=(0.0022)(Vf1)

An arrow is shot at a 20 degree angle with a launch velocity of 78 m/s what is the time aloft?

Taloft=I 2Vy/ag I

Taloft=I 2Vy/ag I

Taloft=| 2Vy/ag |

20

78 m/s

Vy

Taloft= ?

Vy= ?

ag= 9.8 m/s^2

sin(20)=Vy/78

78sin(20)=Vy

26.678 m/s=Vy

Taloft= ?

Vy= 26.678 m/s

ag= 9.8 m/s^2

Taloft=| 2(26.678)/(9.8) |

Taloft= 5.444 s

Elastic Collision

Work

Inelastic Collision

Time Aloft

Kinetic to Potential

W=Fd

**Review**

**Potential to Kinetic Energy**

Inelastic and Elastic Collisions

Time Aloft

Work

Inelastic and Elastic Collisions

Time Aloft

Work

1/2 MaV^2=1/2Fx

M1V1+M2V2=(M1+M2)Vf

M1V1+M2V2=M1Vf1+M2Vf2

Taloft=| 2Vy/ag |

-29.091 m/s = Vf1

An arrow, is drawn back in a recurve bow a distance of 70 cm if the work done is 77 Joules what is the force exerted on the arrow?

70 cm = 0.7 m

W=77 Joules

F= ?

d= 0.7 m

77=F(0.7)

110 N= F