Tuesday, November 24, 2009

Fox 193

Inspired from a question submitted by Ajit.

10 comments:

  1. A new question
    What is the min length of that parabola?

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  2. A very good one!
    But let's solve this one first :)

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  3. Answer is E
    http://i50.tinypic.com/10xy6ur.gif

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  4. Let’s assume that the vel. of the projectile is V at the top left hand corner of the wall as it passes from left to right.. For the most efficient throw V must make angle of 45 deg to the horizontal & this gives us: 20 =(V^2)sin(2 α)/g or V = (20g)^(1/2). Now Vx=(20g)^(1/2)*cos(45) =(10g)^(1/2). Also Vy = (10g)^(1/2).
    Vx remains unaffected by gravity. Now if U be the initial vel. of throw at an angle θ, we can say that Uy^2=Vy^2+2*g*(10)=20g+10g= 30g or Uy =(30g)^(1/2) while Ux =Vx=(10g)^(1/2). This gives us: U^2 = Ux^2 +Uy^2 =30g +10g = 40g or U=(40g)^(1/2) and further, tan(θ) =Uy/Ux =(3)^(1/2) or θ is 60 deg.
    Ajit

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  5. Two great solutions! Thank you guys.
    You have not checked if this is the shortest trajectory, have you?

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  6. I have done some work by trial and error for my math isn't strong enough to directly find the shortest path. I found that at θ=75 deg. i.e. the projectile is chucked from a point~1.92 m from the wall, we get the following eqns. of motion:10=3.732x-Bx^2 and 10=3.732(x+20)-B(x+20)^2 which when solved give us x=3.032 and B=0.1432 so that y=3.732x-0.1432x^2. Now one may find dy/dx and inetgrate (1+(dy/dx)^2)^(1/2) from 0 to 26.064 to get s = 57.388 m. This, I found to be the shortest parabolic path but naturally it will require a larger initial velocity of V(52.13g). Perhaps one can check further if a still shorter path is possible. I leave it to someone else for I've had enough of banana chucking for the moment!
    Ajit

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  7. Thank you Ajit,
    I'll keep that as an open question.
    Great work with bananas!!!
    Thanks again.

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  8. Just a small correction. At θ=75 deg. the stone needs to be chucked from a point 3.023 m from the wall and it lands exactly at the same distance on the side i.e at a total distance of 20.046 m from the origin. I've erroneously mentioned 1.93 m in my earlier post.
    Ajit

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  9. Just too many typos in my post. I'll make just one more attempt to see if I can get it right.
    At θ=75 deg. the banana needs to be chucked from a point 3.023 m from the wall and it lands exactly at the same distance on the other side i.e at a total distance of 26.046 m from the origin i.e the point fom which it is chucked.

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