• Nov 18th 2009, 12:42 PM
PrincessIsrael
Lunar Gravity

On the moon, the acceleration due to gravity is -1.6 m/(sec^2). A stone is dropped from a cliff on the moon and hits the surface of the moon 20 seconds later. How far did it fall? What was its velocity at impact?

So far i have s(20)=0
• Nov 18th 2009, 01:05 PM
Scott H
Because acceleration is defined as the rate of increase of velocity, and velocity the rate of increase of distance, we know that

$\displaystyle \frac{d}{dt}\frac{d}{dt}s(t)=\frac{d^2s}{dt^2}=-1.6.$

To find velocity from acceleration, we integrate:

$\displaystyle \frac{ds}{dt}=\int \frac{d^2 s}{dt^2}\,dt=-1.6t+C.$

As the stone is dropped rather than thrown, its velocity at $\displaystyle t=0$ must be $\displaystyle 0$, and therefore $\displaystyle C$ must be $\displaystyle 0$:

$\displaystyle \frac{ds}{dt}=-1.6t.$

Now all that remains is to perform the second integration

$\displaystyle s(t)=\int \frac{ds}{dt}\,dt=\int -1.6t\,dt,$

and find the value of $\displaystyle C$ for which $\displaystyle s(20)=0$.