# Need help understanding

1. Sep 26, 2005

### F.B

Ok now i am on acceleration due to gravity questions and i am kind of stuck.
Anyways the question is below

Heres what i did.

d=Vit + 1/2(a)(t^2)
d= -2.1(3.8) + 1/2(9.8)(3.8)^2

I think this how you get the answer which is 63 m but i dont get why we are using 3.8 s because at 3.8 s the displacement or distance should be 0 so can anyone please help me out

Sorry im am such an idiot i did the wrong steps the actual question is in the last post

Last edited: Sep 26, 2005
2. Sep 26, 2005

### HallsofIvy

Is there something missing here? Exactly what was the question???

Okay, I guess the question was "when will the balloon be at height 15 meters!" and the missing part was that the balloon was punctured (this is also assuming that air resistance plays no part which not very realistic for even a punctured balloon. Maybe the missing part was "suddenly the atmosphere disappears"!

Your equation is -4.9t2+ 15t- 15= 0 which you can solve by the quadratic formula:
$$x= \frac{-15 + \sqrt{(15)^2-4(-4.9)(15)}{2(-4.9)}$$
Did you forget the the "-" on the -4.9? Certainly -4(-4.9)(15) will be positive and (15)2-4(-4.9)(15)= (15)2+ 4(4.9)(15) will be positive.

Last edited by a moderator: Sep 26, 2005
3. Sep 26, 2005

### F.B

Ohh sorry i forgot the question.

A hot air balloon is moving with a velocity of 2.1 m/s [UP] when the balloonist drops a ballast (a large mass used for height control) over the edge. The ballast hits the ground 3.8 s later.

How high was the balloon when the ballast was released??

Last edited: Sep 26, 2005