## Tuesday, November 24, 2009

### Fox 193

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

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

3. Seems hard:)

4. http://i50.tinypic.com/10xy6ur.gif

5. 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

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

7. 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

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

9. 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

10. 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.