Finding acceleration with only height and time.

[undergradphys]

So people have been arguing about a video games new hero, saying the gravity with which corpses fly off at is too low, and it looks like "moon physics". One guy went so far as to solve for acceleration for the bodies in the video.

"http://www.youtube.com/watch?v=ldMOdxKs8S4

At 11:06 you can see that zombie went up and over a ~10 m tree in ~3 seconds.

a=Voy^2/(2*h)

t/2=v/a

So, 8*h/t^2=a

Thus,
8*10/(3^2) = a = 8.9 m/s^2 ~ equal to EARTH (Especially if you consider drag)


In fact, to get the moons gravity of 1.624 m/s^2 the tree would need to be,

8*h/(3^2) = 1.624 => h = 1.827 meters tall, which is likely SHORTER than out hero."

Im confused where he got his equations, because it doesn't seem valid taht you can calculate acceleration, which is change in velocity, with only the change in distance over time.

 

[kuruman]

Im confused where he got his equations, because it doesn't seem valid taht you can calculate acceleration, which is change in velocity, with only the change in distance over time.

You can because what you have is maximum distance, $h_{max}$ and the time of flight $t_f$ which is also the time required for the initial velocity $v_i$ to reverse direction. Here is how he got his equations. Note that $a$ stands for the magnitude of the acceleration.
From the velocity equation $v_f=v_i-at$ you get,$$-v_i=v_i-at_f~\rightarrow~v_i=\frac{at_f}{2}$$Using the equation relating acceleration, displacement $d$ and initial and final speed, $2(-a)d=v_f^2-v_i^2$ with $d=h_{max}$ and $v_f=0$, $$2(-a)h_{max}=0^2-v_i^2=-\left(\frac{at_f}{2}\right)^2$$from which$$a=\frac{8 h_{max}}{t^2}.$$