Finding g from maximum projectile range on unknown planet

  • Thread starter Thread starter k2var2002
  • Start date Start date
  • Tags Tags
    Magnitude Velocity
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 4K views
k2var2002
Messages
15
Reaction score
0
Your spaceship lands on an unknown planet. To determine the local value of g , you ask a steel-toed crew member to kick a stone, and you find that if she kicks it at 18.0 m/s at various angles, the maximum range she can achieve is 34.8 m .
What is g (m/s)? time (s)? and height (m)?

I decided to use the knowledge of velocity given. Vox= Vocos[tex]\theta[/tex]
I also knew that at 45° maximum distance may be achieved. However, I never did a problem with these unknowns in my physics class so I feel lost as to how to solve it. Any suggestions?
 
Physics news on Phys.org
I have worked through some problem solving methods that I think I need to go through to get to the next steps. I know that I need to solve for acceleration but I'm unsure as to how to do that with the equation y= yo+voyt+1/2at². Once I get the acceleration I believe I can just divide that figure by 9.8m/s² to get the value of g on the planet. Any suggestions are more than appreciated.

Thanks a ton!
 
Hi Kyle! :smile:
k2var2002 said:
I know that I need to solve for acceleration but I'm unsure as to how to do that with the equation y= yo+voyt+1/2at². Once I get the acceleration I believe I can just divide that figure by 9.8m/s² to get the value of g on the planet.

uhh? :confused:

the acceleration is the value of g on the planet. :wink:

Find it (using 45º) by finding t for which y = 0, then find the a for which x = 34.8 at that value of t. :smile:
 
g= Vo^2sin(2a)/R ? Does that make sense for an equation finding g?
 
k2var2002 said:
g= Vo^2sin(2a)/R ? Does that make sense for an equation finding g?

Do you mean g = v02sin2θ/R? How did you get that?

(btw, remember sin45º = 1/√2)