Finding g on unknown planet given only total time and a random height

[tswvyimlis]

Homework Statement

Been working on this for the past hour or so and seem to be going in circles; finally decided to call in the cavalry:

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.

Homework Equations

The Attempt at a Solution

 

[gneill]

The cavalry can't help you without first seeing your attempt at a solution! What formulas are you tinkering with?

 

[tswvyimlis]

Sorry, my first attempt was flawed because I mistakenly equated Vavg.= (Vfinal+Vinitial)/2 and from that eventually came up with the equation 4t*deltaY*(2Vfinal-Vavg)+Vavg=Vfinal and used the reference point of t=1.5s and y=11m to solve for Vfinal. That (probably obviously) didn't work so now I've used the equation y=Vi+Vi*t+(1/2)at² after I plugged in the same reference point I got it down to a=(11-2.5Vi)/(1.125).

I can't seem to get past here. Thanks!

 

[davo789]

I agree, this one takes a bit of thinking and some trial and error!

My hint would be that you need to use $s=ut+\frac{1}{2}at^2$ (1) and $s=vt-\frac{1}{2}at^2$ (2). Using these, you can get an expression for u in terms of a, and then put back into (1) to finally get an answer.

 

[rwisz]

To calculate g on an unknown planet, we can use the equation g = (2h/t^2), where h is the height and t is the total time. In this case, h = 11.0 m and t = 5.00 s - 2.00 s = 3.00 s. Plugging in these values, we get g = (2*11.0 m)/(3.00 s)^2 = 4.89 m/s^2. This is the magnitude of g at the surface of the unknown planet.