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    Tried this question several times and I keep getting the same-wrong- answers for part a Name:  ImageUploadedByStudent Room1456998467.167749.jpg
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    (Original post by drandy76)
    Tried this question several times and I keep getting the same-wrong- answers for part a Name:  ImageUploadedByStudent Room1456998467.167749.jpg
Views: 83
Size:  153.3 KB


    Posted from TSR Mobile
    What do you have so far?

    We know that, horizontally: 30 = Vt\cos 25^{\circ} and that vertically we have: s = 3, u = V\sin 25^{\circ}, a=-g, t =t

    So: \displaystyle 3 = Vt\sin 25^{\circ} - \frac{gt^2}{2}, you know that t = \frac{30}{V\cos 25^{\circ}}

    which, after plugging into the equation should let you solve for V.
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    (Original post by Zacken)
    What do you have so far?

    We know that, horizontally: 30 = Vt\cos 25^{\circ} and that vertically we have: s = 3, u = V\sin 25^{\circ}, a=-g, t =t

    So: \displaystyle 3 = Vt\sin 25^{\circ} - \frac{gt^2}{2}, you know that t = \frac{30}{V\cos 25^{\circ}}

    which, after plugging into the equation should let you solve for V.
    Ive been using Vt= 30/cos(25) and subbing the value given into the y equation to solve for t, and hence V, which i got as 15.6, i also tried using the equation of trajectory and got the same answer
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    (Original post by drandy76)
    Ive been using Vt= 30/cos(25) and subbing the value given into the y equation to solve for t, and hence V, which i got as 15.6, i also tried using the equation of trajectory and got the same answer
    Would the answer happen to be something close to 22.1 ish?
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    (Original post by Zacken)
    Would the answer happen to be something close to 22.1 ish?
    yeah thats the answer
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    (Original post by drandy76)
    yeah thats the answer
    Mind posting your working so I can see where you're going wrong?
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    (Original post by Zacken)
    Mind posting your working so I can see where you're going wrong?
    Name:  ImageUploadedByStudent Room1456999545.651741.jpg
Views: 49
Size:  139.1 KB this is where I'm at ATM
    Name:  ImageUploadedByStudent Room1456999603.654479.jpg
Views: 40
Size:  134.5 KB and this was my previous attempt


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    (Original post by drandy76)
    Name:  ImageUploadedByStudent Room1456999545.651741.jpg
Views: 49
Size:  139.1 KB this is where I'm at ATM
    and this was my previous attempt


    Posted from TSR Mobile
    Remember that your s=ut + \cdots has ut = v\sin \25^{\circ} \times \frac{1}{v \cos 25^{\circ}} = \tan 25^{\circ}, there's no \frac{1}{v} there since the v's cancel.
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    (Original post by drandy76)
    Name:  ImageUploadedByStudent Room1456999603.654479.jpg
Views: 40
Size:  134.5 KB and this was my previous attempt


    Posted from TSR Mobile
    This is all correct up till \frac{1}{2}gt^2 = 10.98 then multiply both sides by 2 and divide by g so t^2 \approx \frac{21.96}{9.8} \approx 2.24 whilst you've got 4 for some reason.
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    (Original post by Zacken)
    Remember that your s=ut + \cdots has ut = v\sin \25^{\circ} \times \frac{1}{v \cos 25^{\circ}} = \tan 25^{\circ}, there's no \frac{1}{v} there since the v's cancel.
    Thanks i finally got it:^_^:, it started to look similar to one of (many) failed working, after looking back i didnt multiply 3 by v^2 initially :doh:
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    (Original post by drandy76)
    Thanks i finally got it:^_^:, it started to look similar to one of (many) failed working, after looking back i didnt multiply 3 by v^2 initially :doh:
    Awesome. See my reply to why your other method with finding t didn't work either.
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    apparently i decided it was appropriate to divide by 4.9 even after multiplying both sides by 2, Mechanics is preventing from even doing basic math
    (Original post by Zacken)
    Awesome. See my reply to why your other method with finding t didn't work either.
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    (Original post by drandy76)
    apparently i decided it was appropriate to divide by 4.9 even after multiplying both sides by 2, Mechanics is preventing from even doing basic math
    Happens to all of us, glad it's sorted now. :-)
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    I saw Zacken doing some Mechanics!!
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    (Original post by TeeEm)
    I saw Zacken doing some Mechanics!!
    Never again. :ninja:
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    (Original post by Zacken)
    Never again. :ninja:
    you should try more often
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    (Original post by TeeEm)
    you should try more often
    I agree, I like mechanics, I'm just not very good at it, I'll work on it sometime
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    (Original post by Zacken)
    I agree, I like mechanics, I'm just not very good at it, I'll work on it sometime
    less time here ...!
    (mind you if you end up at Warwick you will not be doing any .... Just Pure and Stats)
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    (Original post by Zacken)
    I agree, I like mechanics, I'm just not very good at it, I'll work on it sometime
    I love how TeeEm keeps insinuating that you're going to Warwick - such great banter :rofl:
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    (Original post by TeeEm)
    I saw Zacken doing some Mechanics!!
    Good games yesterday....
 
 
 
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