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    Sammy Becht's Avatar
    Sammy Becht Posts: 1, Reputation: 1
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    #1

    Oct 11, 2009, 11:49 AM
    Projectile 8.0 times the speed at maximum height, calculate elevation angle at launch
    The launching speed of a certain projectile is 8.0 times the spped it has at its maximum height. Calculate the elevation angle at launching.
    printz0r's Avatar
    printz0r Posts: 2, Reputation: 0
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    #2

    Oct 20, 2009, 03:07 PM
    The speed of the projectile can be broken into two velocity components. Vy and Vx. At maximum height Vy is equal to 0. The velocity vector V=(Vy^2+Vx^2)^1/2

    We also know, that the initial speed of the projectile is 8 times the amount of max height.

    The initial velocity of the projectile is Vo, and that can be broken into two components as well. Vox and Voy. From vectors, Vo=(Vox^2+Voy^2)^1/2

    Vo=8 Vx... substitute we get... Vox^2 + Voy^2 = 64*Vx^2.

    Tan(x)=Voy/Vox... Voy= Tan(x)*Vox.

    Substitute again.

    Vox^2 + (Tan(x)*Vox)^2 = 64*Vx^2.

    Now one final equation. Vx=Vox because gravity does not act in the X direction but only the y.

    So we divide by Vox^2 and we get

    1+ (Tan(x))^2=64
    marijoi's Avatar
    marijoi Posts: 1, Reputation: 1
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    #3

    Aug 15, 2010, 06:28 AM

    The height h, in meters, of a projectile t seconds, after it is fired from a cannon is given by h=kt-1/2gt^2. Suppose we assume that k= 50 and g= 9.8 m/s^2, when is the projectile at a height of 20 meters?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Aug 16, 2010, 08:54 AM

    You should have started your own thread marijoi.

    use the equation given to you.



    Use the values that were given to you, h = 20, k = 50, g = 9.8.



    You now get a quadratic. Rearrange, and then solve this quadratic.
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