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    Anyone understand how to do question 1b 6 and 17 from the heinemann book.

    I got the correct answer for 6, but my direction of motion was something like 45 degrees, which was completely different to the books answer. So if you have done this question before, could you please tell me if you got a answer similar to my.

    Question 17 is the main one I got stuck on and is as follows:
    A stone is projected from point O on a cliff with a speed of 20m/s at an angle of elevation of 30 degrees.T seconds later the angle of depression of the stone from O is 45 degrees.Find the value of T.

    I have been trying alot of different ideas, but I just dont fully understand it enough to answer it.
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    (Original post by Goodfella1989)
    Anyone understand how to do question 1b 6 and 17 from the heinemann book.

    I got the correct answer for 6, but my direction of motion was something like 45 degrees, which was completely different to the books answer. So if you have done this question before, could you please tell me if you got a answer similar to my.

    Question 17 is the main one I got stuck on and is as follows:
    A stone is projected from point O on a cliff with a speed of 20m/s at an angle of elevation of 30 degrees.T seconds later the angle of depression of the stone from O is 45 degrees.Find the value of T.

    I have been trying alot of different ideas, but I just dont fully understand it enough to answer it.
    For question 17, the question is saying that after T seconds if you drew a line joining the position of the stone to the point from which the stone was thrown, then the angle between that line and the horizontal will be 45degrees{below the horizontal}.
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    What is the books answer for q 17?
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    Its 5.59

    I remember doing this Q: It was one I actually managed to do by myself!!

    If you want help, ask!
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     \\

{resolve upwards} \ u=20\sin 30 \ s=s \ t=t \ a=-g \\

{resolve right} \ u=20\cos 30 \ s=s \ t=t \ a=0

    solve this
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    If u solve upwards u= 20sin 30= 10m/s , we know a=-g but we don't know anything else.
    If you solve downwards we know u= 20 cos30, and thats all.

    I don't understand how the depression helps in this question, unless it can help to find the final velocity.
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    (Original post by Goodfella1989)
    I don't understand how the depression helps in this question, unless it can help to find the final velocity.
    You know the angle of depression of the ball is 45 degrees. \tan 45^\circ = 1. This implies that the horizontal distance moved by the ball is exactly the same as the distance it has fallen below the initial point of release.

    Find expressions for these two distances using the resolved components of the initial velocity and equate them. Solve the resulting equation for T.
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    (Original post by Theorist)
    You know the angle of depression of the ball is 45 degrees. \tan 45^\circ = 1. This implies that the horizontal distance moved by the ball is exactly the same as the distance it has fallen below the initial point of release.
    Sorry I am a little confused could you explain how does tan45=1 ,imply that the horizontal distance that the ball is exactly the same as the distance it has fallen below the initial point of release.
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    Just tried the method:

    S=20cos30T
    - S=20sin30T+0.5*(-9.8)T^2

    0=20cos30T-20sin30T +4.9T^2

    0=((20cos30-20sin30)+4.9T)T

    Therfore
    20sin30-20cos30=4.9T
    T=-1.49s
    Which is not the right answer, so im guessing I must of done something wrong or the method is wrong...
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    Goodfella: this is what I mean.


    I'll work through the question from the beginning. When t = 0, u_y = 20 \sin 30^\circ = 10 and u_x = 20 \cos 30^\circ = 10\sqrt3.

    When t = T, we know that the ball is at an angle of 45 degrees below the initial point of release. Using some trig, \tan 45^\circ = \frac{|y_T|}{|x_T|}. But we also know that \tan 45^\circ = 1. Therefore \frac{|y_T|}{|x_T|} = 1 \Rightarrow |x_T| = |y_T|.

    Now, resolving horizontally, there is no acceleration. So |x_T| = 10\sqrt3 T. I take up to be positive when resolving vertically. Therefore y_T will be negative because the ball is closer to the ground at time T than when it was when t = 0. In other words, -|y_T| = y_T. Remember, |...| denotes the magnitude or positive numerical value of something. So writing the equation for the vertical motion, remembering that g is negative due to my sign convention:

    y_T = u_y T + \frac1 2 a_y T^2
    -|y_T| = 10T - \frac1 2 gT^2
    |y_T| = \frac1 2 gT^2 - 10T.

    But |x_T| = |y_T|. Therefore

    \frac1 2gT^2 - 10T = 10T\sqrt3
    \frac1 2gT^2 - 10(1+\sqrt3)T = 0
    T\left(\frac1 2gT - 10(1+\sqrt3)\right)=0.

    But, physically speaking, we know that T\neq 0, which implies that \frac1 2gT - 10(1+\sqrt3)=0. Rearranging this for T gives T = \frac{20(1+\sqrt3)}{g}, which is the required answer.

    Do you understand now?
    Post any questions you have.
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    ye I understand now and the method you told me was right, I just got my sign in the wrong place, thank you very much for the help
    just 1 more question, is Yt the distance of the vertical component right...
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    [QUOTE=Theorist]
    I take up to be positive when resolving vertically. Therefore y_T will be negative because the ball is closer to the ground at time T than when it was when t = 0. In other words, -|y_T| = y_T.
    QUOTE]

    Another question, how do you know the ball is closer to the ground at time T, is that just an assumption made because the ball has an angle of depression.
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    after trying to do this question recently I still don't understand why take vertical equation as negative
 
 
 
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