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Projectile Motion Help

  1. May 17, 2005 #1
    I have this practice problem that I juts cannot figure out.

    I have my guesses, but I really don't know if I am doing this right.

    This is the diagram we are given:
    Diagram

    And I am supposed to fill out this table:
    Table

    So they tell us that the speed is 24 m/s. And the horizontal distance traveled is 62.340m.

    So my question is, in order to fill out the table, should I use the distance (62.34m) and divide it by the various times in the table to find the new speed for each table row?

    I know the formula for finding Vv is vsin0 (0 being the angle provided).

    So should my formula be: (62.340m/3.00s)sin(-30) for example on the last row of the table?

    Any help would be greatly appreciated.
     
  2. jcsd
  3. May 17, 2005 #2

    OlderDan

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    My interpretation of what you are being asked to do is find the verticle component of the total velocity at various points along the path of the projectile. 24 m/s is the initial velocity, and the direction of the total velocity is given at the different times listed in the table. The label of the velocity at the top of the parabola is confusing, since there is no vertical velocity at that point. I suspect the subscript should have been an H for horizontal. In addition to the given direction, you should know the horizontal component of the velocity at each time. From those two values you can find the vertical component.

    I don't know what other information you may have been given, but this is not a projectile at the surface of the earth. The vertical acceleration has a value other than g.
     
  4. May 17, 2005 #3
    Maybe, but I think the question is asking you to find the velocity of the projectile after a certain time, based upon the iniitial angles given on the Left of the table.

    This is not too hard to do, just manipulate the projectile equations.


    -Ben
     
  5. May 17, 2005 #4
    I re-did the table, using 24 as the set V.

    And got 12 m/s, 8.62 m/s, etc.... 0 m/s, etc..., -8.62 m/s, -12 m/s.

    This seems to make the most sense to me as something traveling upwards will have the same speed when traveling down, correct?
     
  6. May 17, 2005 #5

    OlderDan

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    Your observation about the symmetry is correct, but if 8.62 is your result for the vertical velocity at t = .250 or at .500 then it is not right. Your resolution of the 24 m/sec at 30 degrees into a vertical component of 12 m/sec is consistent with other information in the problem. In particular, the horizontal velocity would be (24 m/sec) cos 30 = 20.78 m/sec, and a 3 second time of flight would give the 62.34 m range as stated.

    When the velocity is at 25.70 degrees at t = .250, the horizontal component must still be 20.78 and you have 20.78 = v_total cos 25.70. Solving that for v_total gives you v_total = 23.06. Then v_v = 23.06 sin 25.7 = 10 m/sec.

    There is a more direct way to do this. You never have to calculate the total velocity. What you have to recognize is that the vertical component divided by the horizontal component is the tangent of the angle. Since the horizontal component is constant, you have

    v_v = (20.78 m/s) tan (theta)

    For 25.7 degrees v_v = 10 m/sec
    For 21.06 degrees v_v = 8 m/sec

    etc

    Now, from another perspective, think about the change in vertical velocity during the time the object is rising. It starts out at 12 m/sec, and when it gets to the top the vertical velocity is zero. That happens in 1.5 seconds. What does that tell you about the vertical acceleration? For one thing, it tells me this projectile is not at the surface of the earth. It is somewhere in a lower gravity environment. From the velocity change in 1.5 seconds you can calculate the acceleration, and once you have it you can calculate the vertical velocity at any time from

    v_v = 12 m/sec + at (a is negative)

    Those calculations must be consistent with the calculations based on the angle (they are for all the ones I checked).
     
  7. May 19, 2005 #6
    Thankyou! That seems to make much more senese now!

    So the gravitational acceleration should be -8m/s^2 then, right?

    And the starting at 12 m/s, and ending at -12 m/s, that would make the average vertical velocity 0 m/s, wouldn't it?
     
  8. May 19, 2005 #7
    Also, if the distance is 62.340m (given) then the time taken to hit the ground will be:

    t= d_H/v_H2 = 62.34/(20.78[2]) = 1.5 s, so is that just the time taken to rise, so it should be doubled in order to find the total time of 3 seconds?

    Or is 20.78 m/s actually supposed to be 10.39 m/s?
     
  9. May 19, 2005 #8

    OlderDan

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    Yes. The average vertical velocity would be zero. The change in vertical velocity would be

    -12 m/s - 12 m/3 = -24 m/s,

    and since that happens in 3 seconds, the acceleration is

    (-24 m/s)/(3s) = -8m/s^2 as you have found.
     
  10. May 19, 2005 #9

    OlderDan

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    Look at this again. The horizontal velocity is constant for the whole 3 seconds. The total time is the full distance of 62.34m divided by the constant velocity 20.78 m/s. In 1.5 second, the time the projectile is rising (or falling), only half of the horzontal distance is covered, so your equation is correct. Half the distance divided by the constant velocity gives you half the time. The velocity is not cut in half; it was the distance you cut in half to get half the time.
     
  11. May 19, 2005 #10
    Thankyou once again, I would be so lost, this booklet seems contradictory to me.
     
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