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    SOOO Confusing!!!



    When resolving when do you know which to use between Sin or Cos :s

    I went on examsolutions and I know how to do the question but I get confused between whether to use Sin or Cos
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    Resolve EVERYTHING. I find that drawing axes on the diagram helps me to decide whether to use sin or cos
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    When you are Stretching and angle, use Sin. When you are Crushing an angle, use Cos.
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    (Original post by I'mBadAtMaths)
    When you are Stretching and angle, use Sin. When you are Crushing an angle, use Cos.
    Thanx !!!
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    can some tell me when to use sin and cos becuase "When you are Stretching and angle, use Sin. When you are Crushing an angle, use Cos." didnt help me much.
    help please
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    please help me !
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    (Original post by mekymonkey)
    can some tell me when to use sin and cos becuase "When you are Stretching and angle, use Sin. When you are Crushing an angle, use Cos." didnt help me much.
    help please
    First off, draw a right-angled triangle. For the question given here, you can draw a dotted line horizontally across from P to the downward vertical line (which goes up to O).

    Now you have your right-angled triangle, label the hypotenuse, the adjacent side and the opposite side in relation to the 20 degree angle. Then using SOHCAHTOA or however you have learnt it, so you can work out whether to use sin or cos.

    With practice you will not need to go through the same drawn out method as you will immediately be able to identify whether you are after the component of force opposite (so use sin) or adjacent (next to it meaning you use cos).

    So for this question, the vertical component of tension is Tcos(20) and the horizontal component of tension is Tsin(20). Then just resolve! Hope this helps you to understand-it took me ages to grasp the concept, but once I was explained it properly I understood how it works
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    lol sorry it took so long to reply, but yh thank you i totally get it now. got m1 next Wednesday!
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    could someone show me how to do a question?

    A ball is thrown vertically upwards with speed u from a point P at height h metres above the ground. The ball hits the ground 0.75 s later. The speed of the ball immediately before it hits the ground is 6.45ms-1. the ball is modeled as a particle.

    a. show that u=0.9ms-1
    b. Find the height above P to which the ball rises before it starts to fall towards the ground.
    c. Find the value of h.
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    (Original post by mekymonkey)
    A ball is thrown vertically upwards with speed u from a point P at height h metres above the ground. The ball hits the ground 0.75 s later. The speed of the ball immediately before it hits the ground is 6.45ms-1. the ball is modeled as a particle.

    a. show that u=0.9ms-1
    b. Find the height above P to which the ball rises before it starts to fall towards the ground.
    c. Find the value of h.
    a. show that u=0.9ms-1

    For this part; we already know everything we need, its all told in the question.
    So we can SUVAT the ball on the way up and on the way down. The time it takes on the way up plus the time it takes to hit the floor from its peak height, will give 0.75s. (told in the question).

    SUVAT for ball on the way up
    S= not needed.
    U= u (told in question)
    V= 0 (peak height)
    A= -9.8 (gravity)
    T=

    Using v=u+at
    0=u-9.8t
    9.8t=u
    t=u/9.8 -------- (1)

    SUVAT for ball on the way down from peak height
    S= not needed
    U= 0 (peak height)
    V= 6.45 (told in question)
    A= 9.8 (gravity)
    T=

    Using v=u+at
    6.45=9.8t
    t = 6.45/9.8 = 0.658163.... (i prefer to keep it in fraction) ---------(2)

    We know that (1) + (2) = 0.75s
    therefore, (1) = 0.75 - (2)
    u/9.8 = 0.75 - 0.658163...
    u/9.8 = 0.09
    u = 0.09 x 9.8
    u = 0.9m/s

    b. Find the height above P to which the ball rises before it starts to fall towards the ground.

    SUVAT for ball on the way up (with knowledge that u=0.9m/s)
    S=
    U=0.9
    V=0
    A=-9.8
    T=

    Using v^2=u^2 + 2as
    0=0.81 - 19.6s
    19.6s = 0.81
    s=0.81/19.6
    s=0.04m

    c. Find the value of h.

    SUVAT the ball going down (gives the entire height the ball reaches up to from the floor)
    S=
    U=0 (peak)
    V=6.45 (from question)
    A=9.8
    T=

    using v^2=u^2+2as
    41.6025=19.6s
    s=2.12257...

    As h = s - 0.04
    h=2.08m


    Hope these are the correct answers and they helped ^^" The way i did it was right though, I think :rolleyes:
 
 
 
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