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        riddles >> hard >> THE EASIEST THROW
        
(Message started by: pcbouhid on Dec 1^st, 2005, 10:38am)

Title: THE EASIEST THROW
Post by pcbouhid on Dec 1^st, 2005, 10:38am

Paul (P) won a historic competition in his school, throwing a ball clear over the roof shown in the picture (10m high, 16m wide), with the least effort.

How far (x meters) away from the wall did he take his stand? (assume the ball is thrown from 2m above the floor).

Title: Re: THE EASIEST THROW
Post by towr on Dec 1^st, 2005, 10:49am

What is the air resistance?

Title: Re: THE EASIEST THROW
Post by Three Hands on Dec 1^st, 2005, 4:11pm

Just as an initial effort, assuming no air resistance:

[hide] If we opt for a 45degree average of throwing, then he should stand 6m away (found by constructing a double-sized 3-4-5 triangle, with the hypotenuese running between P (assuming point P to be 2m from the floor) and the top-right corner of the school, and the right-angle being the line of the vertical wall and perpendicular to this, also passing through point P). This is almost certainly not the correct answer, though ::)[/hide]

Editted for clarity...

Title: Re: THE EASIEST THROW
Post by Eigenray on Dec 1^st, 2005, 7:00pm

I'll take a shot: [hide]8(sqrt(3)-1)[/hide]?

Title: Re: THE EASIEST THROW
Post by Joe Fendel on Dec 1^st, 2005, 7:15pm

x = -16m. ;D

Title: Re: THE EASIEST THROW
Post by pcbouhid on Dec 2^nd, 2005, 5:36am

The original text doesn´t state nothing (or anything?) about "air resistance". In the solution its not considered.

Elgenray, I think you have to explain your shot. Joe doesn´t need. ::)

Note: hope this problem fits in Icarus´"creme-de-la creme". More to come. Soon in this channel. ;D

Title: Re: THE EASIEST THROW
Post by JocK on Dec 2^nd, 2005, 2:15pm

Let's say the ball reaches a height h above the roof. This requires a total energy E = m g (h + 8) + m v²/2. Here m = mass, g = gravitational acceleration, and v = horizontal velocity.

The time t the ball spends above the roof satisfies g t² = h. So: t = sqrt(h/g).

Also, v t = 16, so: v = 16/t = 16 sqrt(g/h).

Hence: E = m g (h + 8) + 128 m g/h = m g (8 + h + 128/h)

E is minimal when h = 8 sqrt(2).

Using a coordinate system centered at the midpoint of the roof, the parabola describing the trajectory of the ball is:

z = 8 sqrt(2) (1 - (x/8)²)

The ball is thrown from a 'height' -8, the corresponding x coordinate follows from:

-8 = 8 sqrt(2) (1 - (x/8)²)

=> x = 8 sqrt(1 + 1/sqrt(2)) = 10.4525 m

Hence, the distance from the wall (x - 8) is 2.45 m.

(Surprisingly close... might have made a calculational error.. but pretty sure about the methodology..! :D )

Title: Re: THE EASIEST THROW
Post by pcbouhid on Dec 3^rd, 2005, 4:15am

Do you agree, Joe? Or are you going to defend your solution? ;D

Title: Re: THE EASIEST THROW
Post by Grimbal on Dec 3^rd, 2005, 12:53pm

I'd say the optimal throw has the ball going through both corners of the building.

The energy necessary to throw the ball is equal to the energy necessary to lift the ball to the corner and throw it from there to the other corner. You only need to optimize that throw. For throwing the ball from one corner to the other, the optimal throw starts at 45°. That lifts the ball 1/4 of 16 m or 4m above the center of the building.

From the apex to the corner it is 8m horizontally and 4m vertically. From the apex to the thrower it is 3 times as far vertically, (8m down from the corner), so it is sqrt(3) times as far horizontally. So, the horizontal distance from the apex is 8m*sqrt(3), and from the building it is 8m·(sqrt(3)-1).

Title: Re: THE EASIEST THROW
Post by Grimbal on Dec 3^rd, 2005, 1:00pm

on 12/02/05 at 14:15:49, JocK wrote:

The time t the ball spends above the roof satisfies g t² = h. So: t = sqrt(h/g).

It has to go up and down...

Title: Re: THE EASIEST THROW
Post by JocK on Dec 3^rd, 2005, 1:49pm

on 12/03/05 at 13:00:58, Grimbal wrote:

It has to go up and down...

Yes, that's why it says:

g t²

rather than:

g t²/2

Yet, I feel that I might have made some obvious error somewher... ???

Title: Re: THE EASIEST THROW
Post by Grimbal on Dec 3^rd, 2005, 3:52pm

You should compute h for half of the time:
h = g*(t/2)²/2

Title: Re: THE EASIEST THROW
Post by JocK on Dec 4^th, 2005, 1:37am

on 12/03/05 at 15:52:57, Grimbal wrote:

You should compute h for half of the time:
h = g*(t/2)²/2

You're absolutely right. That changes my calculation into: E = m g ( 8 + h + 16/h) => h = 4 => x - 8 = 8 (sqrt(3) - 1).

We now agree. Thanks!

Title: Re: THE EASIEST THROW
Post by pcbouhid on Dec 5^th, 2005, 8:12am

We have a problem with the right answer.

Yours are 5.8564...m, and I have 5.856m.

Something is wrong. ;D