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Math Help - Projectiles horizontally

  1. #1
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    Projectiles horizontally

    Question A is ok but try it anyway.I cannot do (B) i need help iv exam tomorrow.no one seems to be able to do this question.

    (A)A projectile is launched horizontally out to sea with a initial velocity of 98m/s from the top of a vertical cliff 78.4 m high.

    (i)what is the time taken to reach the sea?

    (ii)how far from the base of the cliff will it hit the sea?


    (B)I CAN NOT DO NEED IT DONE ASAP PLEASE.

    the angle of the launch is now raised to 30 degrees above the horizontal.


    (i)what is the time taken to reach the sea?
    (ii)how far from the base of the cliff will it hit the sea?
    (iii)how long will it take to reach its greatest height above the point of launch?
    (iv)what is the greatest height above the point of launch?[IMG]file:///C:/DOCUME%7E1/CATHAL%7E1.000/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/DOCUME%7E1/CATHAL%7E1.000/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/DOCUME%7E1/CATHAL%7E1.000/LOCALS%7E1/Temp/moz-screenshot-1.jpg[/IMG]
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  2. #2
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    Quote Originally Posted by Cathaloc2 View Post
    Question A is ok but try it anyway.I cannot do (B) i need help iv exam tomorrow.no one seems to be able to do this question.

    (A)A projectile is launched horizontally out to sea with a initial velocity of 98m/s from the top of a vertical cliff 78.4 m high.

    (i)what is the time taken to reach the sea?

    (ii)how far from the base of the cliff will it hit the sea?


    (B)I CAN NOT DO NEED IT DONE ASAP PLEASE.

    the angle of the launch is now raised to 30 degrees above the horizontal.


    (i)what is the time taken to reach the sea?
    (ii)how far from the base of the cliff will it hit the sea?
    (iii)how long will it take to reach its greatest height above the point of launch?
    (iv)what is the greatest height above the point of launch?
    I will explain the first two

    For (i) you need to remember that the vertical component of the projectile is accelerating under 'g'. So h = u\text{(vertical)}t + \frac12 gt^2

    Carefully fix the sign of u, that is see if it is against g or not.

    For (ii) remember that while your projectile is falling down it has constant horizontal velocity. So if the from the previous step you get time 't' as the time it takes to fall down. Then the distance from the base of the cliff is
    u\text{(horizontal)}t
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  3. #3
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    i really need this done (Part B) step by step so i know exactly how to do it and to ensure im right.
    thanks for your help
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  4. #4
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    Hello, Cathaloc2!

    (A) A projectile is launched horizontally out to sea
    with an initial velocity of 98m/s from the top of a vertical cliff 78.4 m high.

    (B) The angle of the launch is now raised to 30 above the horizontal.
    You need the projectile formulas:

    . . x \;=\; (v_o\cos\theta)t
    . . y \;=\; h_o + (v_o\sin\theta)t - 4.9t^2

    where: . \begin{array}{ccc}h_o &=& \text{initial height} \\ v_o &=& \text{initial speed} \\ \theta &=& \text{angle of elevation} \end{array}


    We are given: . h_o = 78.4,\quad v_o = 98,\quad\theta = 30^o

    The equations are:

    . . x \:=\:(98\cos30^o)t \quad\Rightarrow\quad {\color{blue} x(t) \:=\:49\sqrt{3}\,t}

    . . y \:=\:78.4 + (98\sin30^o)t - 4.9t^2 \quad \Rightarrow\quad{\color{blue}y(t) \:=\:78.4 + 49t - 4.9t^2}



    (i) What is the time taken to reach the sea?
    When is y = 0 ?

    We have: . 78.4 + 49t - 4.9t^2 \:=\:0\quad\Rightarrow\quad 4.9t^2 - 49t - 78.4 \:=\:0

    Quadratic Formula: . t \;=\;\frac{49 \pm\sqrt{49^2 - 4(4.9)(-78.4)}}{2(4.9)} \;=\;\frac{49 \pm\sqrt{3937.64}}{9.8}

    The positive root is: .  t \:\approx\:\boxed{11.4 \text{ seconds}}



    (ii) How far from the base of the cliff will it hit the sea?
    x(11.4) \;=\;49\sqrt{3}(11.4) \;\approx\;\boxed{967.8\text{ m}}


    (iii) How long will it take to reach its greatest height above the point of launch?
    The height function, y \:=\:-4.9t^2 + 49t + 78.4, is a down-opening parabola.
    . . Its maximum is at its vertex: . v \,=\,\frac{-b}{2a}

    We have: . t \:=\:\frac{-49}{2(-4.9)} \:=\:\boxed{5\text{ seconds}}



    (iv) What is the greatest height above the point of launch?
    y(5) \;=\;78.4 + 49(5) - 4.9(5^2) \;=\;\boxed{200.9\text{ m}}

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  5. #5
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    thanks you very much
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