Student Question

# A stone is dropped from the top of a building 960ft high.  What is its impact velocity?

`v = -g t`

so

`t = -v/g`

And we know that

`s = -1/2g t^2`

so `s = -1/2g(-v/g)^2`

solving for v we get

so `v = sqrt(-2gs)`

Since s = -960 ft.

`v = sqrt((960"ft")(2(32.2"ft/sec"^2))) = sqrt(61824"ft"^2"/sec"^2) = 248 "ft/sec" `

`v~~ 250 "ft/sec"`

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Since you give no size of the stone or any other details, I'm going to assume that you are neglecting air resistance.

At the surface of the earth, gravity accelerates objects at 9.81 Newtons per Kilogram, or equivalently, 9.81 metres per second squared (ms^2)

As 1' = 0.3048m, the height you are dropping the stone from is 292.608m. Now using the SUVAT equations in SI units, it is possible to calculate all aspects of the stones trajectory.

We have the acceleration, the distance it needs to travel and its starting point. We don't know (and want to know) its velocity after acceleration and we don't care about how long it takes. The SUVAT equation which fits this scenario is

v^2=u^2+2as

where v is the eventual velocity, u is the starting velocity, a is the acceleration and s is the distance it needs to travel. If you're dropping it from rest, u=0. Thus, you can calculate

v=sqrt(2as)=sqrt(2*9.81*292.608)=75.769 ms^-1.

If you need it in feet per second, thats 248.59 ft*s^-1.

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