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    Page 85, Qu. 9.

    How does one show that:

    6 tan^2 A - 30 tan A + 5 = 0?

    Thanks in advance.
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    Could you post the whole question?
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    Yes. Is that 6tan^2(A-30)*tan(A+5) or 6tan^2(A)-30tan(A)+5, or one of the other groupings I cba to type? :p:
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    A golf ball is projected with speed 49 m/s at an angle of elevation A from a point B on the first floor of a golf driving range. Point B is at a height of 3 and 4/15 metres above horizontal ground. The ball strikes the ground at point Q which is a horizontal distance of 98 metres from point B.

    Show that:

    (6 tan^2 A) - (30 tan A) + 5 = 0

    Thank you!
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    Can anyone help with this, please?
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    Any ideas, anyone?
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    (Original post by Snagglepuss)
    A golf ball is projected with speed 49 m/s at an angle of elevation A from a point B on the first floor of a golf driving range. Point B is at a height of 3 and 4/15 metres above horizontal ground. The ball strikes the ground at point Q which is a horizontal distance of 98 metres from point B.

    Show that:

    (6 tan^2 A) - (30 tan A) + 5 = 0

    Thank you!
    Ignore that it's asking you to show that. Just look at the question; what equations and things can you form? The 6tan^2A - 30tanA + 5 = 0 will fall out on its own.

    First thing I'd do is come up with equations for its horizontal and vertical displacements. You have to post your working (or thoughts on the question) for us to give you more help since it's generally agreed upon to not post full solutions.
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    Hullo Swayum,

    I have now spent about an hour tyring to crack this. I am pretty sure I am just failing to spot some obvious point, but I'd appreciate some more help. I am using the usual sin and cos relations to work out expressions for initial vert and horiz speeds, i.e. horiz speed is a constant 49 cos A, and initial vertical speed is 49 sin A. I have tried to use SUVAT formulae to calculate the time it takes the ball to hit the ground, but can't as I know neither the intial vertical speed nor the terminal vertical speed, nor time t.

    I'd very much appreciate some pointers from you or anyone. Many thanks.
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    Anyone feeling helpful, please?
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    Let t_1 be the time at B. Let t_2 be the time at Q. Then 49t_1\cos A + 98 = 49t_2\cos A \Rightarrow \cos A (t_2 - t_1) = 2. Now find an expression from vertical motion, and eliminate t_1 , t_2 from it using the relation (t_2 - t_1) = \frac{2}{\cos A}
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    Thank you, Kolya, but I have a total blank on this. I have been battering away at it for nearly two weeks now on and off, inluding two hours today. I am sure two minutes with a tutor would clear it up, but I just can't keep wasting any more time on it as I am way behind. What takes a moment to explain in the classroom can never happen at all if one just doesn't get something. Thanks again.
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    I misread the question earlier. Anyway, if you want some closure:

    Horizontally, we have:

    v = 49cosA
    s = 98

    Therefore time when ball is at Q is t = 2/cosA.

    Vertically, for ball travelling from B to Q:

    u = 49sinA
    s = -49/15
    t = 2/cosA
    a = -g = -9.8

    Therefore, using s = ut - 0.5gt^2

    -49/15 = 98tanA - 9.8*2*sec^2A =>
    -1/30 = tanA - 1/5sec^2A =>
    -1 = 30tanA - 6sec^2A =>
    -1 = 30tanA - 6tan^2A - 6 =>
    6tan^2A - 30tanA + 5 = 0. \square
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    Very much appreciated, Kolya. I was thrown today, I think, by your error in reading the question at first (but of course I don't at all blame you for that).

    Blasted trig ident at the root of it. I utterly hate them, and I think was unconsiously kidding myself that there was another way to tackle it without using the wretches.

    Thanks again. Mi casa es tu casa and so on and so forth.
 
 
 
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