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    A golfer strikes a ball with speed 60ms-1. The ball lands in a bunker at the same level, 210 metres away. Calculate possible angles of projection.

    My working:
    S(y)=0......... ......S(x)=210
    u=60sin \theta........ ......u=60cos \theta
    v=-60sin \theta............ t=?
    a=-9.8 ..............t=\frac{210}{60cos \theta}
    t=?

    Substituting t into S=ut+0.5at^2
    S=ut+\frac{1}{2}at^2

    0=\dfrac{(210)60sin \theta}{60cos\theta}-\dfrac{4.9(210)^2}{60^2cos^2 \theta}

    210tan \theta-\dfrac{4.9(210)^2}{60^2cos^2 \theta}=0

    210tan \theta-\dfrac{4.9(210)^2 (1+tan^2 \theta)}{60^2} =0

    (60^2)210tan \theta-4.9(210)^2-4.9(210)^2tan^2 \theta=0

    i then went on and moved the equation to the R.H.S and put my values in the quadratic formula. Did i make a mistake? is this how im supposed to do it?
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    Okay penultimate line, times by cos^2 (and change tan into sin/cos). You can then use one of the double angle identities in reverse so you can get 1 trig function which you can then solve. Your last line is completely wrong btw.
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    Never mind i solved it, Turns out when i moved everything to the R.H.S i got one of the signs wrong( put + instead of -)
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    (Original post by black1blade)
    Okay penultimate line, times by cos^2 (and change tan into sin/cos). You can then use one of the double angle identities in reverse so you can get 1 trig function which you can then solve. Your last line is completely wrong btw.
    I managed to get the answer. Using my last line
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    (Original post by joyoustele)
    I managed to get the answer. Using my last line
    Ohhh I didn't spot you using sec^2=1+tan^2 . I still think my way is easier but of course it works as long as the identities you use are correct .
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    The working for this is a bit easier overall IMHO if you first obtain an expression for t from the vertical motion of the golf ball. (t = ([email protected])/g). If you substitute that “t” in the expression for horizontal motion, you end up with quite a benign trig formula in terms of [email protected]
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    A lovely trick here to find the solution easier is to factorise:

    You have here:

     ut + (1/2) a t^{2} = 0,

     \Rightarrow  t(u + (1/2)at) = 0,

     \therefore t = 0 \quad \text{or} \quad u + (1/2) at = 0.
 
 
 
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Updated: October 23, 2017
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