Kinematics Formulas - Is the textbook right?

[Procrastinate]

Question: A hot-air balloonist is ascending vertically at 1.8m/s during a burn. At the same time, a passenger drops an object from the balloon and notes that the object takes 15s to strike the ground. At what height is the balloon when the object strikes the ground assuming g=0.8m/s/s, no wind resistance to the object's fall, and a steady ascent of the balloon.

I first found out the distance that the object dropped and calculated that to be be 1102.5m. Afterwards, I found out the distance that the balloon rose during that 15s which was 13.5m. I then added these two values together and got 1116m.

However, the answer says that it is 1102.5m but I thought that would have been the distance that the balloon was when the object dropped. Am I wrong or is the textbook wrong?

 

[D H]

The book is right. Show your work so we can help figure out where your mistake is.

 

[clamtrox]

h = gt^2/2 + v₀ t

 

[Procrastinate]

The book is right. Show your work so we can help figure out where your mistake is.

u = o (at rest)
a = 9.8
s = ?
t = 15

s = ut + 1/2at^2

s = 1/2x9.8x15^2

s = 1102.5m

However, I am confused as I was calculating this under the impression that I had only calculated the distance the balloon was away from the ground at the start and I haven't factored in the other part of the equation where I have to calculate the distance of the balloon away from the ground after the object hits the ground.

 

[D H]

u = o (at rest)

The object is initially at rest with respect to the balloon, not the ground.

 

[Jebus_Chris]

The book is right. It asks for the height right when the object is dopped which is 1102.5m.