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    17) A stone is projected from a 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.
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    14) split it into horizontal and vertical components and SUVAT the horizontal and vertical.

    17) draw a diagram but similar to 14 in terms of SUVAT that. just remember that displacement (s) can be negative.
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    (Original post by Zohair09)
    14) split it into horizontal and vertical components and SUVAT the horizontal and vertical.

    17) draw a diagram but similar to 14 in terms of SUVAT that. just remember that displacement (s) can be negative.
    I know all that, but I don't know how to do it.

    14) Horizontally:
    s=4
    a=0
    t=2

    Vertically:
    s=4
    a=-9.8
    t=2

    17)
    Horizontally:
    u=20cos30
    a=0

    Vertically
    a=-9.8
    u=20sin30
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    For question 14 you can see that the angle of projection is 45 degrees, so you can split up the velocity into horizontal and vertical components, I think.

    Oh! and you can use the time and the displacement to find the components of the velocity! :excited:
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    14) You can find the initial horizontal speed by using t=2 and horizontal distance = 4m. The same for the initial verticle speed only that your using the formula s = ut + 1/2at^2, bung in the verticle distance (4m) and time, and rearrange to find u (initial verticle speed).

    Than just do tantheta = verticle speed / horizontal speed, and find theta which will give you your direction of motion.

    The range is quite simple, just find the time it takes for the ball to reach the ground by making s = ut + 1/2at^2 equal to 0, at that time the ball has reached its range (horizontal distance) so jus sub in that t value into s = ut.
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    (Original post by + polarity -)
    For question 14 you can see that the angle of projection is 45 degrees, so you can split up the velocity into horizontal and vertical components, I think.

    Oh! and you can use the time and the displacement to find the components of the velocity! :excited:
    It doesn't match the answer in the back of the book of 4.82m at 80degrees.
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    (Original post by samir12)
    14) You can find the initial horizontal speed by using t=2 and horizontal distance = 4m. The same for the initial verticle speed only that your using the formula s = ut + 1/2at^2, bung in the verticle distance (4m) and time, and rearrange to find u (initial verticle speed).

    Than just do tantheta = verticle speed / horizontal speed, and find theta which will give you your direction of motion.

    The range is quite simple, just find the time it takes for the ball to reach the ground by making s = ut + 1/2at^2 equal to 0, at that time the ball has reached its range (horizontal distance) so jus sub in that t value into s = ut.
    Horizontally:
    4=2u
    u=2

    Vertically:
    4=2u-4.9*4
    4=2u-19.6
    2u=23.6
    u=11.8

    tan feta = 11.8/2

    feta=80.3 degrees.

    I don't understand the next bit though.
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    (Original post by Deep456)
    It doesn't match the answer in the back of the book of 4.82m at 80degrees.
    Oh sorry! I totally forgot to use the equations of motion. Just ignore my posts.
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    Right and 17?
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    (Original post by Lou Reed)
    Split V into horizontal and vertical components and equate.

    Vh = 10root3
    Vv = |10 - 9.8t|

    10root3 = -10 + 9.8t
    t = 0.75 seconds

    I think that's it...what's the answer in the book?

    note: i used modulus on Vv so we get a positive t
    Nope it is not the answer. 5.59 s is the answer.
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    This question is driving me mad.
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    Lou Reed had the right idea, but was trying to work out when the speed was acting at 45 degrees (although I think they made a mistake with the modulus etc because I believe that would be after ~2.79 seconds).

    Angle of depression = angle between horizontal and the line from O to the particle (below horizontal). Drawing a diagram, you should see that when the angle of depression is 45 degrees, (vertical distance) / (horizontal distance) = tan (45). In this case this is relatively easy, because - as tan (45) = 1 - they are the same.

    So find the horizontal distance at time T, the vertical distance (downwards) at time T, and equate them then solve for T.

    Spoiler:
    Show

    Horizontal distance at time T = 20cos(30)T
    Vertical Distance (taking down as downwards positive) = 4.9t^2 - 20sin(30)T


    Note I get 5.58, so I'm guessing your book has a slight rounding error? (unless I've made a mistake somewhere)
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    (Original post by gcseeeman)
    Lou Reed had the right idea, but was trying to work out when the speed was acting at 45 degrees (although I think they made a mistake with the modulus etc because I believe that would be after ~2.79 seconds).

    Angle of depression = angle between horizontal and the line from O to the particle (below horizontal). Drawing a diagram, you should see that when the angle of depression is 45 degrees, (vertical distance) / (horizontal distance) = tan (45). In this case this is relatively easy, because - as tan (45) = 1 - they are the same.

    So find the horizontal distance at time T, the vertical distance (downwards) at time T, and equate them then solve for T.

    Spoiler:
    Show

    Horizontal distance at time T = 20cos(30)T
    Vertical Distance (taking down as downwards positive) = 4.9t^2 - 20sin(30)T


    Note I get 5.58, so I'm guessing your book has a slight rounding error? (unless I've made a mistake somewhere)
    Ah thanks! :yep:
 
 
 
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