I am now doing acceleration due to gravity. And i am confused about something. I'll just write down 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 3.8s later.

 

How high was the balloon when the ballast was released?

 

This is what i did.

 

d=Vit + 1/2at^2

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

 

This is supposed to work but i dont get why we are using 3.8s as the time because at 3.8s the distance/displacement is 0. So how come this works. Can anyone please explain this.

you're simply working out the distance travelled by the mass, from the ballon at the altitude where it droped it (what you wish to find) and the ground. As you know how long this takes it's easy to calculate the distance travelled...

This is supposed to work but i dont get why we are using 3.8s as the time because at 3.8s the distance/displacement is 0. So how come this works. Can anyone please explain this.

 

At t=3.8s the position may be zero (if that's what you define the ground to be), but the displacement is how far it traveled from t=0 until that time. So the displacement is from where it was dropped to the ground.