How do I determine the height from which the stone is dropped?

[morson]

Homework Statement

A balloon is rising vertically at a constant speed of 21 m/s. A stone is dropped from the balloon. The stone strikes the ground 10 seconds later. Assuming that air resistance is negligible and that acceleration on Earth due to gravity is 9.8 m/s2, from what height is the stone dropped?

Homework Equations

v = u + at
dv/dt = a
dx/dt = v

The Attempt at a Solution

This is screwing with my head. It's an obvious displacement problem, but where would you set the origin, to which the displacement is relative?

Here's my attempt:

The acceleration of the stone at any time is 9.8, and because this is constant
acceleration, v = u + at where a = 9.8 and u = 0 (because if you haven't dropped the stone yet, it must have a velocity of 0) .:. v = 9.8t

Integrating that you get x = 9.8(t^2/2) + c .:. x = 4.9t^2 + c

Here's where I run into trouble. WHERE IS THE ORIGIN?? Is it at the ground? At the balloon? At the height from which the stone is dropped? How do I define the displacement?

I tried x = 0 at the ground .:. x = 0 when t = 10 .:. 0 = 4.9(100) + c

.:. c = 490

so x = 4.9t^2 + 490

So the displacement at any time is equal to 4.9t^2 + 490.

We know that the stone is dropped after 0 seconds, so the height from which the stone is dropped would be a height of 490 metres, according to what I came up with.

If I'm wrong, could you please point out where I went wrong? In my approach I completely ignored the fact that the balloon rises at 21m/s. Is that part relevant?

 

[Doc Al]

Taking the orgin to be at a height of zero is perfectly OK. Ignoring the initial speed of the stone is not OK--it's not zero! (What's the initial speed of the stone?)

Choose a coordinate system and sign convention, such as "up is positive; down is negative". Note that the acceleration would be -9.8 m/s^2.

 

[morson]

Taking the orgin to be at a height of zero is perfectly OK. Ignoring the initial speed of the stone is not OK--it's not zero! (What's the initial speed of the stone?)

Choose a coordinate system and sign convention, such as "up is positive; down is negative". Note that the acceleration would be -9.8 m/s^2.

Thanks a lot. I'll revise what I did, then:

The initial speed of the stone is 21m/s, if I'm re-evaluating correctly. The acceleration is -9.8m/s^2.

The velocity is -9.8t + c ==> at t = 0, v = 21 ==> 21 = c so:

v = -9.8t + 21

Integrate to get X (displacement) = -9.8(t^2/2) + 21t + c
= -4.9t^2 + 21t + c

x = 0 when t = 10 ==> 0 = -490 + 210 + c ==> c = 280

x = -4.9t^2 + 21t + 280

At x = h, t = 0: h = 280

280 metres?

If it's correct, thanks a lot. If it's not, thanks anyway.

 

[Doc Al]

Looks good.

 

[morson]

Okay, awesome.