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If I'm given angle of projection and the range how do I work out Initial velocity? Watch

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    I have a question where a particle is projected at 45 degrees to the horizontal and it lands 100m away. I'm trying to figure out how I'd calculate the initial velocity from this (u) but my brain cells are firing blanks. I'd appreciate any help.

    Obviously the component of trajectory are
    Horizontal: Ucos45
    Vertical: Usin45

    If I try to work out for example the time of flight by working in the vertical direction and I take v=0 I have t=(v-u)/a but that leaves me with two unknowns(the value of t and the value of u). I'm thinking maybe I have to substitute the above formula for t into another formula, but I'm not sure. :confused::confused::confused:
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    (Original post by Wunderbarr)
    The range is the maximum horizontal distance travelled by the particle.

    You have been given an angle, so can you use the fact that speed = distance / time and rearrange where necessary, as well as trigonometry ?

    Also, to find the time you will need to look at the vertical motion only, and the particle's journey under the influence of gravity.

    Hopefully this is enough information to point you in the right direction, if not then please ask
    I know all of that, but I still can't seem to get anywhere.
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    (Original post by student1856)
    I know all of that, but I still can't seem to get anywhere.
    Ok, so to work out the time of flight you should be using the formula:
    s = ut + 0.5at2
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    (Original post by Wunderbarr)
    Ok, so to work out the time of flight you should be using the formula:
    s = ut + 0.5at2
    You don't have the value for the vertical component of s, you don't have the time of flight and you don't have the initial velocity...
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    (Original post by student1856)
    You don't have the value for the vertical component of s, you don't have the time of flight and you don't have the initial velocity...
    For s, you can always look at where it is at the beginning and the end of the journey. Since s means displacement, the beginning and end have the same displacement relative to the ground, 0.

    For u, replace it with what you know is the vertical component of the initial velocity.

    You're trying to work out t here so you will get something in terms of u.

    So in summary, s = 0. u = usin(45) and t = ?

    Work out t in terms of u and you should be able to take it from there.
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    I apologize if this is wrong but you know the horizontal component so why not ucos45=100m ? Horizontal velocity doesn't change.
    P.S I haven't studied in about a month so my brain also out of the game.


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    (Original post by OnTheHorizon)
    I apologize if this is wrong but you know the horizontal component so why not ucos45=100m ? Horizontal velocity doesn't change.
    P.S I haven't studied in about a month so my brain also out of the game.


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    It's S=UT, so 100m=ucos45xtime of flight. And we don't know the time of flight.
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    Sorry


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    We have to consider the horizontal and vertical components separately. Since it lands 100m away, we also know the time taken for both the horizontal and vertical componenets has to be the same as well.

    Vertical:
    Spoiler:
    Show
    Here we use the suvat equations to find an expression for t:

     u = u\sin45

a = -g

s = 0

t = ?

    s = ut +\frac{1}{2}at^2

0 = u\sin45t -\frac{1}{2}gt^2

t(u\sin45 -\frac{1}{2}gt) = 0

    t = 0, t =\dfrac{2u\sin45}{g}
    Horizontal:
    Spoiler:
    Show
    We can now substitute our expression for t into our horizontal component:

     u = u\cos45

s = 100

t =\dfrac{2u\sin45}{g}

     s = ut

100 =\dfrac{2u^2\cos45\sin45}{g}

    Then just rearrange and solve for u.
    Hope this helps
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    (Original post by Wunderbarr)
    For s, you can always look at where it is at the beginning and the end of the journey. Since s means displacement, the beginning and end have the same displacement relative to the ground, 0.

    For u, replace it with what you know is the vertical component of the initial velocity.

    You're trying to work out t here so you will get something in terms of u.

    So in summary, s = 0. u = usin(45) and t = ?

    Work out t in terms of u and you should be able to take it from there.
    So i do 0= usin45t + 0.5(-9.8)t^2
    4.9t^2=usin45t =======>divide both sides by t

    4.9t=usin45

    t=usin45/4.9...I have two unknowns.
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    (Original post by NeverLucky)
    We have to consider the horizontal and vertical components separately. Since it lands 100m away, we also know the time taken for both the horizontal and vertical componenets has to be the same as well.

    Vertical:
    Spoiler:
    Show
    Here we use the suvat equations to find an expression for t:

     u = u\sin45

a = -g

s = 0

t = ?

    s = ut +\frac{1}{2}at^2

0 = u\sin45t -\frac{1}{2}gt^2

t(u\sin45 -\frac{1}{2}gt) = 0

    t = 0, t =\dfrac{2u\sin45}{g}
    Horizontal:
    Spoiler:
    Show
    We can now substitute our expression for t into our horizontal component:

     u = u\cos45

s = 100

t =\dfrac{2u\sin45}{g}

     s = ut

100 =\dfrac{2u^2\cos45\sin45}{g}

    Then just rearrange and solve for u.
    Hope this helps
    Thank you so much. I got the correct answer by applying this method. Did you learn this during M2 as a formula that was taught to you or did you derive it on your own? Thanks anyway.
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    (Original post by student1856)
    Thank you so much. I got the correct answer by applying this method. Did you learn this during M2 as a formula that was taught to you or did you derive it on your own? Thanks anyway.
    You do learn projectile motion as part of M2 but the method for these types of questions is pretty much always the same. The theory for projectile motion is quite simple and you rely on a few basic facts like the time of motion is the same for both the horizontal and vertical components. If you're doing M2, then once you do a few of these questions, you'll know what to do immediately.
 
 
 
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