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Homework Help: Projo with angle

  1. Jan 1, 2012 #1
    1. a projectile is fired into the air at a projection angle of 30° above the ground. its initial speed is 327 m/s. what is its speed 3 seconds later

    2. vf^2=Vi^2+2AΔX

    3. So, the angle is 30, Vi=327m/s, Tf=3. trying to find Vf....I'm guessing combine horizontal and vertical, but I have no clue. Help please!
  2. jcsd
  3. Jan 1, 2012 #2


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    What are the initial horizontal and vertical components of velocity? What is the vertical acceleration? Is there any horizontal acceleration? Do you know a kinematic equation that relates final velocity, initial velocity, acceleration and time (because this would be far more appropriate)?
  4. Jan 1, 2012 #3
    I don't know what the horizonatl and vertical components of velocity are....can you help please!

    Vertical acceleration is 9.8m/s^2 (I think)

    No horizontal acceleration


    Then where does the 30° come in?
  5. Jan 1, 2012 #4


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    the 30 degrees is the angle of elevation. imagine you have a cannon you raise it up 30 degrees above the horizontal and you shoot. the cannon ball starts out with an initial velocity of Vi and the moment it leaves gravity is pulling it down at 9.8 m/s^2 so you use the 30 degrees to divide the Vi into two component velocities Vx and Vy using your knowledge of trigonometry. The Vy is affected by the gravity but the Vx isn't.
  6. Jan 1, 2012 #5


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    As Jedishrfu has advised, use trig (sin and cos) to resolve the initial velocity into horizontal and vertical components. Surely, you would've covered this in coursework.

    The magnitude of vertical acceleration is [itex]g[/itex] as you have stated, but don't forget the direction. The usual convention is to depict "up" as positive and "down" as negative. So if the initial vertical velocity [itex]u_y[/itex] is [itex]+1 ms^{-1}[/itex] (just an example), then the constant vertical acceleration is [itex]-9.8 ms^{-2}[/itex].

    The equation for acceleration you quoted is essentially correct, but the usual form it's written in is [itex]v = u + at[/itex], where v is the final velocity (your Vf), u is the initial velocity (your Vi), a is the acceleration (your A) and t is the time interval (your difference of Tf minus Ti).

    Correct, there is no horizontal acceleration (because you can neglect wind resistance, etc.) So the horizontal velocity [itex]u_x[/itex] remains constant.

    Now find [itex]u_x[/itex] and [itex]u_y[/itex] (the initial horizontal and vertical components of the velocity, respectively) with trig. Then figure out what happens to the vertical velocity after 3 seconds using that equation. Then, once you've got the final vertical velocity [itex]v_y[/itex], figure out how to add that back to the horizontal velocity (which doesn't change) to get the magnitude of the final velocity (which is the final speed). Remember, vector addition is not as simple as just adding two quantities. Hint: Pythagoras.
    Last edited: Jan 1, 2012
  7. Jan 1, 2012 #6
    Okay this is EXACTLY where I am getting stuck! Could you work this out for me?
  8. Jan 1, 2012 #7


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    Unfortunately, forum rules prevent others from showing you the work word for word. However, we can guide you.

    Draw a right angled triangle. The hypotenuse (slanted side) represents the initial velocity [itex]u[/itex] (at a 30 degree angle to the horizontal). The two perpendicular sides represent [itex]u_x[/itex] and [itex]u_y[/itex]. Can you now write down the trig. ratios (sine and cosine) of the 30 degree angle in relation to the ratios [itex]\frac{u_x}{u}[/itex] and [itex]\frac{u_y}{u}[/itex]?
  9. Jan 2, 2012 #8


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    Hi Let It Be, your class should have (and probably has) covered vectors and vector components before going into projectile motion.

    Here is an example which you can apply to this problem: An object has a velocity of 50 m/s at an angle of 60 degrees above the horizontal. Find the horizontal (or x) and vertical (or y) components of the velocity.

    So the vector "50 m/s, at 60 degrees above the horizontal" is represented by this figure:


    And the components -- that is, the horizontal and vertical sides of the right triangle formed by the 60-degree vector -- are worked out using the definitions of sine and cosine:

    cos = (adjacent leg) / hypotenuse

    sin = (opposite leg) / hypotenuse​


    Hope that helps.

    p.s. these figures are from http://www.physicsclassroom.com/class/vectors/u3l2d.cfm
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