- #1

- 35

- 0

## 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?