1. acceleration/gravity problem

You land on an unknown planet somewhere in the universe that clearly has weaker gravity than Earth. To measure g on this planet you do the following experiment: A ball is thrown upward from the ground. It passes a windowsill 11.0 m above ground and is seen to pass by the same windowsill 2.00 s after it went by on its way up. It reaches the ground again 5.00 s after it was thrown. Calculate the magnitude of g (the acceleration due to gravity) at the surface of this planet.

2. Originally Posted by thedoge
You land on an unknown planet somewhere in the universe that clearly has weaker gravity than Earth. To measure g on this planet you do the following experiment: A ball is thrown upward from the ground. It passes a windowsill 11.0 m above ground and is seen to pass by the same windowsill 2.00 s after it went by on its way up. It reaches the ground again 5.00 s after it was thrown. Calculate the magnitude of g (the acceleration due to gravity) at the surface of this planet.
The distance of the ball above the ground as a function of time after its is thrown is:

$\displaystyle s=-\frac{g\,t^2}{2}+v_0\,t$

we have $\displaystyle s=0 \mbox{ m}$ at $\displaystyle t=0, \ 5 \mbox{ sec}$, and $\displaystyle s=11 \mbox{ m}$ at $\displaystyle t=t_1, \ t_1+2 \mbox{ sec}$.

These tell us that

$\displaystyle v_0=(5/2) \,g$,

and so the roots of (as these are the times at which the ball passes the window):

$\displaystyle 11=\frac{g\,t^2}{2}+(5/2)g\,t$

differ by 2. Which with a bit of jiggery-pokery (manipulation of the quadratic formula mainly) tells us that:

$\displaystyle 25-\frac{88}{g}=4$

which we can solve for $\displaystyle g$, to find that $\displaystyle g=88/21$

RonL

3. Can you go into more depth on the quadratic methods you used to get the second to last equation? You lost me there.

4. Originally Posted by CaptainBlack
and so the roots of (as these are the times at which the ball passes the window):

$\displaystyle 11=\frac{g\,t^2}{2}+(5/2)g\,t$

differ by 2. Which with a bit of jiggery-pokery (manipulation of the quadratic formula mainly) tells us that:
Originally Posted by thedoge
Can you go into more depth on the quadratic methods you used to get the second to last equation? You lost me there.
Yes.

Rearrange the quadratic by multiplying through by $\displaystyle 2/g$ to get:

$\displaystyle t^2+5t-22/g=0$

The roots of this are:

$\displaystyle t=\frac{-5\pm \sqrt{25-(4\times 22)/g}}{2}$

These differ by:

$\displaystyle \left[ \sqrt{25-88/g}\right]=2$.

Square:

$\displaystyle 25-88/g=4$.

so:

$\displaystyle g=88/21$

RonL