Solving the Air Balloon & Pellet Problem

[Kandy]

A hot air balloon is ascending straight up at a constant speed of 7.0 m/s. When the balloon is 12.0m above the ground, a gun fires a pellet straight up from the ground level with an initial speed of 30.0 m/s. Along the paths of the balloon and the pellet, there are two places where each of them has the same altitude at the same time. How far above the ground level are these places?
For this problem, I need to find two distances, so I tried to find the two times for the distances first but I can't work it out. This is what I have:
let v=velocity of balloon, t=time, V=velocity initial, a=acceleration
vt=Vt+0.5(-9.81)t^2
7t=30t+(-4.91)t^2
7t=30t-4.91t^2
0=23t-4.91t^2
then I factored it
0=t(23-4.91t)
t=0, 4.68 or 4.7
I'm missing one more time and distance or is this time also incorrect? I don't know how to solve this problem, please help me out

 

[lightgrav]

you need to decide WHEN you're going to start timing ...
either when the balloon leaves the ground,
or when the bullet leaves the ground.

I would start the timer when the pellet is fired.
Where is the balloon when MY stopwatch reads zero?

 

[Kandy]

I redid the problem but I still can't get it because the two times are the same. What did I do wrong now?

vt-12.0=Vt+0.5(-9.81)t^2
7.0t-12.0=30.0t+(-4.91)t^2
7.0t-12.0=30.0t-4.91t^2
0=-4.91t^2+30.0t+12.0

P=-58.9
S=30.0

(-4.91t^2+31.85t)(-1.85t+12.0)
t(-4.91t+31.85)(-1.85t+12.0)

t=0

-4.91t+31.85=0
-4.91t=-31.85
t=6.49

-1.85t+12.0=0
-1.85t=-12.0
t=6.49

 

[daniel_i_l]

1) why did you write vt - 12? When the bullet is fired the balloon is 12m over the ground.
2) The initial speed of the bullet is 30 + 7 (which is the speed of the balloon) - unless your using some other referance system.
3) Don't forget to add the bullets initial position to it's position equation.

 

[Kandy]

i don't understand why the speed of the balloon needs to be added to the initial speed of the bullet.

 

[lightgrav]

Kandy,

you should have 12[m] + 7[m/s] t = 30[m/s] t - ... ,
which SHOULD result in 0 = -12[m] + (30 - 7)[m/s] t ...

your writing would be a lot easier to read
if you started sentences with the SUBJECT

y(balloon) = 12[m] + v_i t
y(pellet) =

Don't try to write one very complicated sentence
that includes all the important information in some hidden form!
Write a few small statements, instead, so you won't get confused.
(and you're allowed to use words to explain).

 

[Kandy]

12.0[m] + 7.0[m/s] t = 30.0[m/s] t - 0.5 (-9.81[m/s^2]) t^2
0 = -4.91[m/s^2] t^2 + 23.0[m/s] - 12.0[m]

factor

product=59 sum=23.0

the two numbers are 2.95 and 20.05

(-4.91 t^2 + 2.95t) (20.05 t - 12.0)
t (4.91 t + 2.95) (20.05 t - 12.0)

t=0

4.91 t + 2.95 = 0
4.91 t = -2.95
t = 0.601

20.05 t - 12.0 = 0
20.0 t = 12.0
t= 0.599

now I substitute the times in for distance

12.0[m] + 7.0[m/s] t
12.0[m] + (7.0[m/s]) (0.601)
12.0[m] + 4.207[m]
= 16.207[m]

12.0[m] + 7.0[m/s] t
12.0[m] + (7.0[m/s]) (0.599)
12.0[m] + 4.193[m]
= 16.193[m]

That's only a difference of 2cm (which makes me feel like it's incorrect) but is this correct. Plz let it be

 

[Fermat]

12.0[m] + 7.0[m/s] t = 30.0[m/s] t - 0.5 (-9.81[m/s^2]) t^2
0 = -4.91[m/s^2] t^2 + 23.0[m/s] - 12.0[m]
...

This is correct.

You should now just use the quadratic formula.

you have,

at² + bt - c = 0

t = {-b +/- sqrt(b² - 4ac)}/2a

You should get t = 0.6s and t = 4s (approx)

 

[Kandy]

I got the same answer (0.6 and 4) now, but I used a graphing calculator; I don't know how to get to t={-b+/- sqrt... because I haven't learned how yet.