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wu :: forums    riddles    hard (Moderators: Eigenray, SMQ, william wu, ThudnBlunder, Grimbal, Icarus, towr)    THE EASIEST THROW « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: THE EASIEST THROW  (Read 1392 times) pcbouhid Uberpuzzler     Gender: Posts: 647 THE EASIEST THROW   The_easiest_throw_drawing.JPG « on: Dec 1st, 2005, 10:38am » Quote Modify 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). IP Logged Don´t follow me, I´m lost too. towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: THE EASIEST THROW   « Reply #1 on: Dec 1st, 2005, 10:49am » Quote Modify What is the air resistance? IP Logged Wikipedia, Google, Mathworld, Integer sequence DB Three Hands Uberpuzzler     Gender: Posts: 715 Re: THE EASIEST THROW   « Reply #2 on: Dec 1st, 2005, 4:11pm » Quote Modify Just as an initial effort, assuming no air resistance:   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   Editted for clarity... « Last Edit: Dec 1st, 2005, 4:12pm by Three Hands » IP Logged Eigenray wu::riddles Moderator Uberpuzzler     Gender: Posts: 1948 Re: THE EASIEST THROW   « Reply #3 on: Dec 1st, 2005, 7:00pm » Quote Modify I'll take a shot: 8(sqrt(3)-1)? IP Logged Joe Fendel Junior Member     Posts: 68 Re: THE EASIEST THROW   « Reply #4 on: Dec 1st, 2005, 7:15pm » Quote Modify x = -16m.   IP Logged pcbouhid Uberpuzzler     Gender: Posts: 647 Re: THE EASIEST THROW   « Reply #5 on: Dec 2nd, 2005, 5:36am » Quote Modify 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. « Last Edit: Dec 2nd, 2005, 5:43am by pcbouhid » IP Logged Don´t follow me, I´m lost too. JocK Uberpuzzler     Gender: Posts: 877 Re: THE EASIEST THROW   « Reply #6 on: Dec 2nd, 2005, 2:15pm » Quote Modify 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 )       « Last Edit: Dec 2nd, 2005, 4:02pm by JocK » IP Logged solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it. xy - y = x5 - y4 - y3 = 20; x>0, y>0. pcbouhid Uberpuzzler     Gender: Posts: 647 Re: THE EASIEST THROW   « Reply #7 on: Dec 3rd, 2005, 4:15am » Quote Modify Do you agree, Joe? Or are you going to defend your solution? IP Logged Don´t follow me, I´m lost too. Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: THE EASIEST THROW   « Reply #8 on: Dec 3rd, 2005, 12:53pm » Quote Modify 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). IP Logged Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: THE EASIEST THROW   « Reply #9 on: Dec 3rd, 2005, 1:00pm » Quote Modify on Dec 2nd, 2005, 2:15pm, 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... IP Logged JocK Uberpuzzler     Gender: Posts: 877 Re: THE EASIEST THROW   « Reply #10 on: Dec 3rd, 2005, 1:49pm » Quote Modify on Dec 3rd, 2005, 1:00pm, 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...  ???                 IP Logged solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it. xy - y = x5 - y4 - y3 = 20; x>0, y>0. Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: THE EASIEST THROW   « Reply #11 on: Dec 3rd, 2005, 3:52pm » Quote Modify You should compute h for half of the time: h = g*(t/2)2/2 IP Logged JocK Uberpuzzler     Gender: Posts: 877 Re: THE EASIEST THROW   « Reply #12 on: Dec 4th, 2005, 1:37am » Quote Modify on Dec 3rd, 2005, 3:52pm, 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!     IP Logged solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it. xy - y = x5 - y4 - y3 = 20; x>0, y>0. pcbouhid Uberpuzzler     Gender: Posts: 647 Re: THE EASIEST THROW   « Reply #13 on: Dec 5th, 2005, 8:12am » Quote Modify We have a problem with the right answer.   Yours are 5.8564...m, and I have 5.856m.   Something is wrong. « Last Edit: Dec 5th, 2005, 8:13am by pcbouhid » IP Logged Don´t follow me, I´m lost too. Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy   - medium => hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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