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Tricky Projectile Motion

  1. Feb 14, 2006 #1
    A projectile is shot from the edge of a cliff h = 265 m above ground level with an initial speed of v0 = 105 m/s at an angle of 37.0° with the horizontal

    Xo=0m Yo=0m
    X= 1361m YTop of Cliff=265m
    Ax= 0m/s/s Ay=-9.8m/s/s
    Vo= 105 m/s Vx= 83.86 m/s
    Theta= 37.0
    T= 16.23 s

    I need to determine the vertical component of the velocity at the point directly above the ground before impact. I calculated the vertical velocity from launch to the maximum Y to be 63.2 m/s.

    I tried calculating this several ways but I know all of the following are wrong:

    v=105+(-9.8)(16.23)= -54
    V^2=105^2+2(-9.8)(0-265)= 127.35
    V^2=0^2+2(-9.8)(0-265)= 72.07

    I know I'm missing something here but I cant quite put my finger on it. Could someone direct me to what ever this is?
     
  2. jcsd
  3. Feb 14, 2006 #2

    berkeman

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    Staff: Mentor

    Given the initial upward velocity, how much time does it take for the projectile to go up, stop, and then fall back down to the ground? How far will the projectile have travelled in that amount of time?
     
  4. Feb 14, 2006 #3
    The maximum hight the projectile reaches above the cliff is 203.7 meters. The total downward distance the projectile must travel is 468.7 meters
     
    Last edited: Feb 14, 2006
  5. Feb 14, 2006 #4
    Total downward distance is 468.7 m
    The total time is 16.23 seconds. It takes the projectile 6.4 seconds to travel up so the total downward time should be 9.83

    Still trying to determine the velocity of the projectile just before it hits the ground.

    0=468.7+V(9.83)+.5(-9.8)(9.83)^2
    V=.49 <---------------------------I know this is still wrong
     
  6. Feb 14, 2006 #5

    berkeman

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    Is there any difference in the vertical component of velocity between this projectile at impact, and a rock dropped from the 468.7m that you calculated?
     
  7. Feb 14, 2006 #6
    V^2=63.19^2+2(-9.8)(0-468.7)
    V=114.8 is also wrong
     
  8. Feb 15, 2006 #7

    berkeman

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    I just did the math, and I don't get 468.7m for the top of the trajectory. Maybe go back and re-check your math for calculating the max y. Your initial Vy is correct, but something else must be off.

    I used the equation for Vy to figure out how long it takes for the vertical motion to go to zero at the top of the arc, and used that time t in the equation for the position y(t) to tell me how high above the top of the cliff the projectile got. Then I used the equation for the position y of the falling object to tell me how long it took to fall all the way to the ground, and plugged that time t into the equation for the vertical velocity Vy to tell me the speed at that final time of impact. I get a max y of a bit over 600m total at the top of the arc, which gives a max Vy at impact on the ground of a bit over 100m/s directed downward. What is the correct answer?
     
  9. Feb 15, 2006 #8
    I don't know why people make a mess of such easy question.
    You should use Energy conservation theorem here.
    Change in potential enrgy = Change in Kinetic energy.
    And now just use the concept that Horizontal mommentum is conserved.
     
  10. Feb 15, 2006 #9

    berkeman

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    Hah! Good point. The KE on the way down past the top of the cliff is the same as on the way up. Doh! Clever shortcut. :biggrin:
     
  11. Jan 19, 2007 #10
    solution to the "particle being projected from 265m cliff" question

    If I interpreted it correctly this is a very straightforward question

    let s = displacement a = acceleration u = initial velocity
    and v = final velocity

    By considering the vertical motion of the particle from cliff to ground:

    a = -9.8 ms-2
    s = -265m
    u = 105sin37 ms-1
    v = ?

    Using the following constant acceleration formula

    v^2 = u^2 + 2as
    v^2 = (105sin37)+ (2 x -9.8
    v^2 = 3993.049 + 5194
    v^2 = 9187.049
    v = -95.8 ms-1 <--- ANSWER
    That's the vertical component of velocity when it reaches the ground (negative because it's going down)

    Hope that helped !
     
    Last edited: Jan 19, 2007
  12. Jan 19, 2007 #11

    cristo

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    Chromium blade: welcome to PF. Please note that it is against PF rules to give out full solutions to homework questions-- we aim to guide the student to the answer with hints, and corrections.

    Since this is about a year old, I don't see there being a problem, but please remember this in future.
     
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