# Homework Help: Calculate Velocity

1. Sep 15, 2007

### Heat

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

A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00 m/s, releases a sandbag at an instant when the balloon is a height h = 40.0 m above the ground After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive.

Compute the position of the sandbag at a time 1.25 s after its release.

2. Relevant equations
Velocity = Initial Velocity + Acceleration(Time)

or

3. The attempt at a solution

From the problem I have been able to obtain the following values for the equation.

a = -9.8m/s^2
V initial = 0 m/s
X initial = 40m
time = 1.25s

so plugging these values in =

Vx = 0 + (-9.8)(1.25)
Vx = -12.25 m/s

Can anyone help me determine where I am making my mistake, or if it's all wrong. :(

2. Sep 15, 2007

### rocomath

how can x initial be 40m it states that the "height h = 40m"?

3. Sep 15, 2007

### Heat

isn't that when x is released at 40m.

so x initial = 0? and x final = 40?

also for a prior problem,

I did use x initial as 40, but time in earlier problem was .165s

so I did x-40.0 = 0 + .5(-.98)(.165)^2
and got 39.8666. ?? Which was right according to solution.

4. Sep 15, 2007

### rocomath

well you're using the equation

$$x=x_{0}+v_{0x}t+\frac{1}{2}a_{x}t^{2}$$

you need to replace the x's with y's so there is no confusion

$$y=y_{0}+v_{0y}t+\frac{1}{2}gt^{2}$$

also, you don't want to assume that it hit the ground, it could still be in free fall. you can calculate when it does hit the ground though by finding when $$v_{y}=0$$

Last edited: Sep 15, 2007
5. Sep 15, 2007

### Heat

ok so it would be

y= 0 + 5m/s) + .5(-9.8)(1.25s)

= -71.56

how can y be (-)?

6. Sep 15, 2007

### rocomath

what is y initial? and what is your equation supposed to be?

what kind of calculator are you using? i just calculated what you gave me and i got nothing close to that.

7. Sep 15, 2007

### Heat

"how can x initial be 40m it states that the "height h = 40m"?"

I believe there was a misunderstanding before, as I was already considering y initial is 40m.

y = 40 + 5(1.25)+.5(-9.8)(1.25)^2
= 38.69375

8. Sep 15, 2007

### rocomath

yes that's what i got, is that correct?

9. Sep 15, 2007

### Heat

yes

Thank you.

10. Sep 15, 2007

### Heat

a quickie, to determine velocity at .165

it would be Vy = V initial + at

= V = 5 + (-9.8)(.165)
the velocity at .165 seconds would be 3.383

does this sound correct?

11. Sep 15, 2007

### rocomath

yes that is what i would do, so if it's wrong we're both wrong :D

but $$v_{y}=v_{0y}+gt$$ is the formula that would give you $$v_{y}$$ is quickest.

12. Sep 15, 2007

### Heat

thank you again, I managed to get that correct , so that means we are both right :)

13. Sep 15, 2007

### rocomath

yes! victory :)

14. Sep 15, 2007

### Heat

another one, this time it tells me to recheck my calculations of sig figs

I need to find : How many seconds after its release will the bag strike the ground?

y-y initial = vinitiaal (time) + .5(g)(t^2)

I got:

-4.9x^2 + 5x + 45 = 0

I did quadratic formula and got

(-5 +- 30.11644069) / (-9.8)

t= -2.562 and t= 3.583

I choose positive time, and says to recheck my calculation or rounding errors sig figs. This only appears when answer is really near.

15. Sep 15, 2007

### rocomath

i would say t = 3.60 s since all your other values have at least 3 sigs

16. Sep 15, 2007

### Heat

Incorrect. :(

I have one more try.

These are the answer I placed.

3.6 ...the program evaluated it as 3.60
3.58 ....the program evaluated it as 3.58
3.583310275 ...the program evaluated it as 3.58

but is the work correct, I tried going over and over, but I get the same number.

17. Sep 15, 2007

### rocomath

you put in your initial height as 45.0m, it should be 40.0m

18. Sep 15, 2007

### Sabellic

I got:

Velocity Initial (hereafter called Vi) = -5.00m/s
Acceleration (hereafter called a)= 9.81m/s^2
Time (hereafter called t)= 1.25s
Displacement (hereafter called s)= x

s=Vi*t + 1/2*a*t^2
x=(-5)(1.25)+(1/2)(9.81)(1.25^2)
x=(-6.25)+7.66
x=1.41

Therefore s=1.41.

In 1.25 seconds after being dropped, the body undergoes a displacement of 1.41m meaning that it is 38.59m above the ground.

19. Sep 15, 2007

### rocomath

well you're substracting from your initial position, but why not go ahead and just compute it all together?

20. Sep 15, 2007

### Heat

well that 45 was the mistake, once I replaced it with 40 it was all good to go. :) thanks to all of you.

21. Sep 15, 2007

### Heat

the last part, I seem to be close to, but I believe I'm doing it the wrong way.

Question asks: What is the greatest height above the ground that the sandbag reaches?

I would logically think that the greatest height above the ground that it reaches is 40m, as that's when it gets released.

The feedback I got this incorrect answer is to recheck calculation, or that I may have made rounding errors or sig figs..

and if I do it algebraically: it would look something like this ::

vx = vo + at

40.0 + (-9.8) (0)
= 40.0