How Do You Calculate Initial Velocity in Archery Physics Problems?

[killaI9BI]

Homework Statement

For archery practice, a knight's squire drops sandbags from a 12.0 meter tower. At exactly the same time the sandbag is dropped, the knight shoots an arrow up at the sandbag from the base of the tower. If the arrow strikes the sandbag at exactly 1.1 seconds, calculate a) how far from the ground the arrow strikes the sandbag, and b) the arrow's initial velocity

Homework Equations

d = ½(vi + vf)t
vi = (d/t) – ((a x t)/2)

The Attempt at a Solution

a) d = ½(-9.8)1.1
d = -5.39

12 - 5.39 = 6.61
The arrow strikes the sandbag at 6.6 m from the ground

b) vi = (6.61/1.1) – (((-9.8) x 1.1)/2)
vi = 6 – (-5.39)
vi = 0.41 m/s

Does this look right?

 

[SteamKing]

Homework Statement

For archery practice, a knight's squire drops sandbags from a 12.0 meter tower. At exactly the same time the sandbag is dropped, the knight shoots an arrow up at the sandbag from the base of the tower. If the arrow strikes the sandbag at exactly 1.1 seconds, calculate a) how far from the ground the arrow strikes the sandbag, and b) the arrow's initial velocity

Homework Equations

d = ½(vi + vf)t
vi = (d/t) – ((a x t)/2)

The Attempt at a Solution

a) d = ½(-9.8)1.1
d = -5.39

12 - 5.39 = 6.61
The arrow strikes the sandbag at 6.6 m from the ground

b) vi = (6.61/1.1) – (((-9.8) x 1.1)/2)
vi = 6 – (-5.39)
vi = 0.41 m/s

Does this look right?

You might want to check this calculation:

a) d = ½(-9.8)1.1
d = -5.39

Remember, the sandbags are free-falling after being dropped, so their velocity is accelerating as they fall.

Don't forget to indicate the correct units for your answers.

 

[killaI9BI]

d = vi X t + 1/2a X t²

that would be a better formula because (-9.8) is the rate of acceleration, not velocity. Thank you for that!

d = 1.1 + 1/2(-9.8) X 1.1²
d = (-4.829)
d = (-4.8) m

12 – 4.829 = 7.171m

a)7.2 m

b)v_i = (d/t) – ((a x t)/2)
v_i = (7.17/1.1) – (((-9.8) x 1.1)/2)
v_i = 11.9 m/s

Does that look right now?

 

[SteamKing]

d = vi X t + 1/2a X t²

that would be a better formula because (-9.8) is the rate of acceleration, not velocity. Thank you for that!

d = 1.1 + 1/2(-9.8) X 1.1²
d = (-4.829)
d = (-4.8) m

12 – 4.829 = 7.171m

a)7.2 m

b)v_i = (d/t) – ((a x t)/2)
v_i = (7.17/1.1) – (((-9.8) x 1.1)/2)
v_i = 11.9 m/s

Does that look right now?

You're veering out of control now. You fixed one problem and have introduced another into your calculations:

d = vi X t + 1/2a X t²

This is the correct formula. So far, so good.

d = 1.1 + 1/2(-9.8) X 1.1²
d = (-4.829)
d = (-4.8) m

12 – 4.829 = 7.171m

What is vi for the sandbags when they are dropped from the tower?

 

[killaI9BI]

I thought that v_i would be zero for the sandbags.

should it be (-9.8) m/s?

 

[Nathanael]

I thought that v_i would be zero for the sandbags.

It would be zero.

d = vi X t + 1/2a X t²
...
d = 1.1 + 1/2(-9.8) X 1.1²

This seems to imply that the initial velocity is 1 m/s (zero times 1.1 is zero, not 1.1)

 

[killaI9BI]

face palm!

d = vi X t + 1/2a X t²
d = ½(-9.8) X 1.1²
d = (-5.929) m

12 – 5.929 = 6.071 m

a) 6.1 m
b) vi = (d/t) – ((a x t)/2)
v_i = (6.071/1.1) – (((-9.8) x 1.1)/2)
v_i = 5.519 – (-5.39)
v_i = 10.9 m/s

Thank you very much!

 

[killaI9BI]

just to be certain, can you confirm that what I've done looks right?

thanks again!

 

[Nathanael]

just to be certain, can you confirm that what I've done looks right?

thanks again!

Yes you're correct, and you're welcome.

 

[killaI9BI]

yay!