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    -Write down the angle needed to project an object on a flat surface so that it achieves maximum horizontal displacement. Ignore air resistance.

    -A projectile is projected with speed v at angle theta to the horizontal. The projectile is on a flat platform raised h metres above the ground, and the ground is flat as well. Derive an expression for theta which gives the maximum horizontal displacement of the projectile. Ignore air resistance.


    http://postimg.org/image/o23uzglcx/
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     sin\theta = \frac {1} {\sqrt { 2 + 2gh/v^2 }}
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    Using the idea that  s = vt we can find an equation to model this situation.

    The vertical displacement will be:  -h

    The vertical acceleration will be:  -g

    The vertical velocity will be:  \nu \sin{\theta}

    We can use the following equation and solve for t, leaving us with all the components we need to define the horizontal displacement since the horizontal component of velocity will be constant throughout the motion.

     s = ut + \frac{1}{2} a t^2

    I'll leave the rest for you.
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    I know the answer lol, I just thought it's a really nice question to share. The first part is meant to be a really easy warm up, it's the second part which is hard.
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    (Original post by L-Tyrosine)
    I know the answer lol, I just thought it's a really nice question to share. The first part is meant to be a really easy warm up, it's the second part which is hard.
    I've done this before, very nice. There's a method to do it with vectors as well, not my style but interesting none the less.
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    (Original post by Zacken)
    I've done this before, very nice. There's a method to do it with vectors as well, not my style but interesting none the less.
    How did you do it without vectors?
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    (Original post by L-Tyrosine)
    How did you do it without vectors?
    Normal resolving and SUVAT fam.
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    (Original post by L-Tyrosine)
    How did you do it without vectors?
    ...
    Attached Images
  1. File Type: pdf Further Projectile_Question 1 and 2_27.04.2015.1.pdf (394.3 KB, 87 views)
  2. File Type: pdf Further Projectile_Question 1 and 2_27.04.2015.2.pdf (433.0 KB, 69 views)
  3. File Type: pdf Further Projectile_Question 1 and 2_27.04.2015.3.pdf (350.0 KB, 65 views)
  4. File Type: pdf Further Projectile_Question 1 and 2_27.04.2015.4.pdf (230.2 KB, 87 views)
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    (Original post by Zacken)
    ...
    Sorry, I should've explained the question better, but the raised surface is flat, as is the ground which is below the raised surface, but the question you did looks interesting as well.

    I added a picture to the original post.
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    (Original post by L-Tyrosine)
    Sorry, I should've explained the question better, but the raised surface is flat, as is the ground which is below the raised surface, but the question you did looks interesting as well.

    I added a picture to the original post.
    I only properly read the question now. Yours is much more trivial, it's not really STEP like to be honest.
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    (Original post by Zacken)
    I only properly read the question now. Yours is much more trivial, it's not really STEP like to be honest.
    Oh ok, did you get the expression? I thought it'd be worthy of at least STEP 1. Try it and you might be surprised
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    (Original post by L-Tyrosine)
    Oh ok, did you get the expression? I thought it'd be worthy of at least STEP 1. Try it and you might be surprised
    4 am where I am, I'll do it in the morning, I'll probably be surprised at how hard I find it!
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    (Original post by Zacken)
    4 am where I am, I'll do it in the morning, I'll probably be surprised at how hard I find it!
    Did the question prove to be harder than you initially thought?
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    (Original post by L-Tyrosine)
    Did the question prove to be harder than you initially thought?
    Forgot to do it, sorry. Just scribbled it onto a piece of paper, is it

    \displaystyle R = \frac{v \cos \theta}{g} \left(v\sin \theta + \sqrt{v^2 \sin^2 \theta + 2gh}\right)

    Then differentiate, set=0, etc...
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    (Original post by Zacken)
    Forgot to do it, sorry. Just scribbled it onto a piece of paper, is it

    \displaystyle R = \frac{v \cos \theta}{g} \left(v\sin \theta + \sqrt{v^2 \sin^2 \theta + 2gh}\right)

    Then differentiate, set=0, etc...
    You run into problems when you set=0 and try to solve, that's where it becomes not so straightforward/you have to think of an alternate approach
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    (Original post by L-Tyrosine)
    Did the question prove to be harder than you initially thought? Added the answer in a spoiler.
    To maximise:

    \displaystyle \theta = \arccos \sqrt{\frac{2gh + v^2}{2gh + 2v^2}}
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    (Original post by Zacken)
    To maximise:

    \displaystyle \theta = \arccos \sqrt{\frac{2gh + v^2}{2gh + 2v^2}}
    Hmm, that answer is equivalent to the one in the spoiler. How did you solve the resulting equation? I guess the question really was easy lol
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    (Original post by L-Tyrosine)
    Hmm, that answer is equivalent to the one in the spoiler. How did you solve the resulting equation? I guess the question really was easy lol
    I can't see your spoiler. I'm going out with my girlfriend, I'll be back in a few hours. Won't be able to reply till then, sorry!
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    STEP maths is such a savage paper, I rate not only those that succeed in it but even the ones who attempt it.
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    (Original post by Zacken)
    I can't see your spoiler. I'm going out with my girlfriend, I'll be back in a few hours. Won't be able to reply till then, sorry!
    Spoiler in the original post, your expression is equivalent though
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    (Original post by L-Tyrosine)
    Spoiler in the original post, your expression is equivalent though
    Another equivalent ones is \displaystyle \theta = \arctan \left(\frac v{\sqrt{v^2+2gh}} \right)
 
 
 
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