# acceleration/gravity problem

• Jan 26th 2007, 05:21 AM
thedoge
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.
• Jan 26th 2007, 05:52 AM
CaptainBlack
Quote:

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:

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

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

These tell us that

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

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

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

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

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

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

RonL
• Jan 26th 2007, 06:01 AM
thedoge
Can you go into more depth on the quadratic methods you used to get the second to last equation? You lost me there.
• Jan 26th 2007, 06:17 AM
CaptainBlack
Quote:

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

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

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

Quote:

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 $2/g$ to get:

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

The roots of this are:

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

These differ by:

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

Square:

$25-88/g=4$.

so:

$g=88/21$

RonL