# Homework Help: Calculating Initial Velocity

1. Aug 24, 2014

### killaI9BI

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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?

2. Aug 24, 2014

### SteamKing

Staff Emeritus
You might want to check this calculation:

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

3. Aug 24, 2014

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

Does that look right now?

4. Aug 24, 2014

### SteamKing

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

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

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

5. Aug 24, 2014

### killaI9BI

I thought that vi would be zero for the sandbags.

should it be (-9.8) m/s?

6. Aug 24, 2014

### Nathanael

It would be zero.

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

7. Aug 24, 2014

### 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)
vi = (6.071/1.1) – (((-9.8) x 1.1)/2)
vi = 5.519 – (-5.39)
vi = 10.9 m/s

Thank you very much!

8. Aug 25, 2014

### killaI9BI

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

thanks again!

9. Aug 25, 2014

### Nathanael

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

10. Aug 25, 2014

yay!