Expert: Alon Mandes - 4/21/2009

Question
If an arrow is shot straight upward on the moon with a velocity of 58 m/s, its height (in meters) after t seconds is given by s(t) = 58 t - 0.83 t^2.

part one had the velocity after 10 seconds to be 41.4 m/s and the second part 69.9 sec for the arrow to return and hit the moon.

My question is, at what velocity in m/s will the arrow hit the moon?

I had been working on this with another expert but he got maxed out so I couldn't get a follow up, but what he had said was...
"It will return with the reverse velocity of what is was shot at.
Note that on earth, we neglect air resistance.  On the moon, we don't even need to worry about air resistance since there is no air.
If you have read this far, I might as well say just take the
negative of how fast it was shot to get the return velocity.
A negative is how fast the arrow is rising.
If you take the negative of this (and get a positive),
the number gotten is fast it is coming down."

and so I tried doing -41.4 and 41.4 but that didn't work so than i tried plugging in -10 for the equation v(t) = 58 - 1.66t which got my my first velocity. I'm not sure how to figure this out. Please help!

Answer
First of all , let's review some facts :
1. The velocity of the arrow as function of time is :
  v(t)=s'(t)=58-1.66t so s'(t=10)=58-16.6=41.4 m/s
2. We want to know how much time takes the arrow to reach it's
  maximum height . So we solve v(t)=0 -> t=58/1.66=34.9 s
  Which mean , the whole journey (tell the arrow hits back the
  ground) will take 2*34.9=69.8 seconds.
3. We are interested in finding , the impact velocity with the
  ground. To do so, we consider the other half symmetric of the
  jurney : the fall ; the arrow start falling from rest , & it's
  velocity will be V(t)=gt. Where g is the gravitation of the
  moon 0.83*2=1.66 . Thus, V(t)=1.66t . The arrow hits the ground
  after 34.9 s . So, V(t=34.9)=34.9*1.66 =~ 58 m/s . The same as
  the lunching velocity.


Alon.