# Am I following this right??

• Sep 27th 2010, 02:50 PM
Am I following this right??
Hi, Stuck on this one, I have a drawing but it does not make sense.
Is this 1D or 2D kinematics??

a hot air balloon is rising straight up with a speed of 3.0 m/s. A ballast bag is released from rest relative to the balloon when it is 9.5 m above the ground.

How much time elapses before the ballast bag hits the ground.
So I have Vo = 0 m/s as it's from rest relative to the balloon?? is that correct.
a = -9.8 m/ss
d = 9.5 m

Do I use d = Vot + 1/2 at^2. And if so how to do I get the 2 t's into 1. I tried to re arrange and it's not working out.

Jo
• Sep 27th 2010, 03:47 PM
mr fantastic
Quote:

Hi, Stuck on this one, I have a drawing but it does not make sense.
Is this 1D or 2D kinematics??

a hot air balloon is rising straight up with a speed of 3.0 m/s. A ballast bag is released from rest relative to the balloon when it is 9.5 m above the ground.

How much time elapses before the ballast bag hits the ground.
So I have Vo = 0 m/s as it's from rest relative to the balloon?? is that correct.
a = -9.8 m/ss
d = 9.5 m

Do I use d = Vot + 1/2 at^2. And if so how to do I get the 2 t's into 1. I tried to re arrange and it's not working out.

Jo

Take down as positive:

u = -3 m/s
d = 9.5 m
a = 9.8 m/s^2
t = ?

Think about these values. In particular, think about the sign they have and think about why u is not equal to zero.

Then use the appropriate formula for uniform straight line motion.
• Sep 27th 2010, 03:55 PM
So the u above is Vo??
Not use to that letter for something, or what do you label it as?
Because it's relative to the balloon, anything going down, that was negative, is now positive then??
• Sep 27th 2010, 04:07 PM
mr fantastic
Quote:

So the u above is Vo??
Not use to that letter for something, or what do you label it as?
Because it's relative to the balloon, anything going down, that was negative, is now positive then??

Yes, my u is your Vo.

If you take up as positive (which is what you seem to want to do) and note that the origin is the balloon at the time of release (I took this as the origin in my previous reply too but assumed you realised this) , then:

u = 3 m/s
d = -9.5 m
a = -9.8 m/s^2
t = ?

Note: It seems you might need to spend more than 8 minutes doing the thinking I suggested that you do.
• Sep 27th 2010, 04:25 PM