AP Physics C Projectile Question

  1. A projectile is fired horizontally from a gun that is 63.0 m above flat ground. The muzzle velocity is 250 m/s.

    (a) How long does the projectile remain in the air? (in seconds)
    I got 3.96 s

    (b) At what horizontal distance from the firing point does it strike the ground? ( in meters) I got 19.44 m

    (c) What is the magnitude of the vertical component of its velocity as it strikes the ground? I got 187 m

    because v(x) is a constant and v sub zero is zero in the vertical direction

    i did 63=.5(9.8)t^2
     
    Last edited: Jun 25, 2006
  2. jcsd
  3. Hootenanny

    Hootenanny 9,678
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    What value are you using for g?
     
  4. 0rthodontist

    0rthodontist 1,253
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    Part a. is correct, the other two parts are not. Where are you having trouble?
     
  5. actually 3.58 s is the time
     
  6. Hootenanny

    Hootenanny 9,678
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    I think he's is off slightly, I get 3.58 seconds.

    Edit: Just seen your revised post now Musicman :smile:

    Perhaps if you show your working for the remaing two questions.
     
  7. mrjeffy321

    mrjeffy321 882
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    I assume you want someone to check you answers?

    For the first part, I used the formula,
    d = -1/2 at^2
    where d is the distance the projectile falls, a is the acceleration due to gravity (-9.81 m/s^2) and t is the time it takes to hit the ground.
    63 m = -1/2 (-9.81 m/s^2) * t^2
    t = sqrt (63 / 4.905) = sqrt (12.844)
    t = 3.58 seconds

    For the 2nd part, I used my answer for time in the first part along with this formula,
    d = v*t
    where d is the horizontal distance traveled, v is the constant horizontal velocity (250 m/s), and t is the time (3.58 s).
    d = 895.96 meters

    For part (c),
    I assumed the horizontal component of the velocity was constant, only the vertical component changed from the intial to final states. To calculate the final vertical velocity, I used the following formula,
    v = a*t
    where v is the velocity after a constant acceleration (9.81 m/s^2 due to gravity) is experienced for a time t (3.58 s),
    v = 9.81 m/s^2 * 3.58 s
    v = 35.12 m/s

    Then I added the two velocity vectors together (horizontal and vertical components) using the Pythagorean theorem to get the final, overall, velocity's magnitude,
    v = 252.45 m/s
     
    Last edited: Jun 25, 2006
  8. Hootenanny

    Hootenanny 9,678
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    Cum on mrjeffy, don't post complete solutions, we can't watch them squirm then! :biggrin:
     
  9. I get 3.58 seconds too.
    Anyway: [tex]t = \sqrt{\frac{128}{g}}[/tex]
     
  10. but for the 2nd part, wouldnt i use the vertical direction equation
    x=(Vo)(t)?
     
  11. Hootenanny

    Hootenanny 9,678
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    The initial vertical velocity is zero. And besides, it is accelerating in the vertical plane (due to gravity). However, you do know that it takes 3.58 seconds to hit the floor. How far horizontally will it have travelled in 3.58 seconds? That is the range, you already know that is travels 63m in the vertical plane. Do you follow?
     
  12. mrjeffy321

    mrjeffy321 882
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    All while I was typing, a bunch of stuff happened, I was just under the impression he wanted to compare his answers (assuming they were correct) with someone else's.

    You initial vertical velocity is zero, so multiplying this by the time will give you zero distance.
    Plus, perpendicular vectors (velocities) are independent....the vertical component of velocity does not [directly] effect the horizontal distance traveled.
     
  13. [tex]x_{max} = v_0t[/tex]

    [tex]x_{max} = 250 * 3.58[/tex]

    [tex]x_{max} = 895 m[/tex]
     
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