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Help with Physics Problem

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Post September 17th, 2008, 9:17 pm

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A golf ball with an initial angle of 32????????? lands exactly 224m down the range on a level course. Neglecting air friction, what initial speed would achieve this result?

Can anyone help me out with this? I feel like it's really simple, but I'm just not seeing it.

Post September 17th, 2008, 9:44 pm

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Post September 17th, 2008, 9:47 pm
jayman Premium Member
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yes.. it's basically the same thing as a gun fire control problem with out a time variable,and windage. hell if i can remember the formula

Post September 17th, 2008, 9:55 pm

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i was thinking somewhere along the lines of:

distance = 1/2 x acceleration x time^2

but that would give us 2 variables. so eh.

Post September 17th, 2008, 10:01 pm
jayman Premium Member
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there's alot missing here.. how high does the golf ball go? how long does it take to get there? with either of these you could plug it into the right formula and get an answer.


Post September 17th, 2008, 10:08 pm

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Post September 17th, 2008, 10:12 pm
jayman Premium Member
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wait.. no thats physics.. there's like decelleration and what kind of arc things make..... and crap.. i have no idea, but it looks like there's enough info to figure it out.. you gotta find the arc it would make to get that distance.. that would be geometry, it would make some sorta parabola thing and land right where they said it did, so make one that starts at 32 degreesand goes where you want it. . then measure 1'2 the distance and the highth of peak of the arc. from there all you gotta do is figure out the velocity it akes to get it half way, gravity is doing the rest


Post September 17th, 2008, 10:14 pm
yoshifreak Premium Member
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Use the range formula: R = [(Vi^2)sin2 (38*)]/g

I believe its around 47m/s?

Post September 17th, 2008, 10:25 pm

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Ok, here's the equation to use and it does not have time which is even better.

distance=(initial velocity)^2 over g times the sine of 2 theta.

or solved for initial velocity:

(initial velocity)=((d*g)/sin2(theta))^(1/2)

The one half power is a square root.

Using this formula and your information I got an answer of 49.415 m/s. This is too many significant figures for the problem, but I would guess that you would know the correct number to use.

Post September 17th, 2008, 10:46 pm

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Post September 17th, 2008, 10:56 pm

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You're welcome. I usually find that solving for the thing you are looking for before putting in numbers can reduce many algebraic errors. The less you have to worry about signs or something like that, the better.


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