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The Physics of Archery

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Joe Draus

on 24 May 2014

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Transcript of The Physics of Archery

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?
(real world data)
(real world data)
(real world data)
(real world data)
(real world data)
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

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
Full transcript