Why use 3.8 s in the kinematic equation if displacement is zero at that time

[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 don't 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 I am am such an idiot i did the wrong steps the actual question is in the last post

 

[HallsofIvy]

Ok now i am on acceleration due to gravity questions and i am kind of stuck.
Anyways i'll show you the question its faster this way.

A hot air balloon is moving with a velocity of 2.1 m/s [UP] when

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

Heres what i did.

d=Vit + 1/2(a)(t^2)
15=15t + 1/2(9.8)(t^2)
15=15t - 4.9t^2
-4.9t^2 +15t - 15 =0

I have to do the quadratic after this but i get a negative inside the roots. Can anyone please tell me why and help me.

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.

 

[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??