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Homework Help: Trouble calculating a projectiles height after a certain distance

  1. Jan 19, 2013 #1
    trouble calculating a projectile's height after a certain distance

    1. The problem statement, all variables and given/known data

    A pebble is fired from a slingshot with a velocity of 30.0m/s. If it is fired at an angle of 30deg to the vertical, what height will it reach? If its fall is interrupted by a vertical wall 12m away, where will it hit the wall in relation to the starting position of the pebble in the slingshot?

    2. Relevant equations

    VF2 = Vo2 + 2ad

    and the 3 other uniform acceleration equations, i guess.

    3. The attempt at a solution

    ok, so i used the above equation to solve for d
    0=[(30)(cos60)m/s]2 + (-9.8m/s)d

    d = 34.44m ~approx

    my trouble comes when i try and calculate how high the pebble will be after 12 meters. if i understand correctly, the question is asking how high up the wall will the pebble hit if it is 12 meters away.

    regardless, i am having difficulty creating an equation to solve for that.

    it may be interesting to note that i dont know whether the ball will have passed 12m by the time it goes 34m up; however, i assume it will be beyond 12m horizontally, so the ball will still be going up when it hits the wall at 12m.

    i greatly appreciate any help.
     
    Last edited: Jan 19, 2013
  2. jcsd
  3. Jan 19, 2013 #2

    haruspex

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    You dropped the 2 in 2ad, but got the right value for d, so I assume that was a transcription error.
    How high will the pebble be at time t? How far will it have travelled horizontally at time t?
     
  4. Jan 19, 2013 #3
    sorry, it was a transcription error.

    i dont understand how to bring time (t) into the mix. it feels like a 2 part solution though. such as D = V*T, and use d=12m, v=horizontal velocity, and solve for t. then after that maybe do another vertical calculation for what is d after the resulting (t) seconds.

    maybe something like that?
     
  5. Jan 19, 2013 #4

    haruspex

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    What other kinematic equations do you know relating distance, speed and time to a constant acceleration?
     
  6. Jan 19, 2013 #5
    i suspect you are hinting towards d=Vf(t)+1/2a(t)2
     
  7. Jan 19, 2013 #6
    i think my confusion here is mostly in regards to translating the X and Y information.

    i dont know if i am working only with vertical or if i need to utilize both X and Y to calculate this height after # seconds.
     
  8. Jan 19, 2013 #7

    haruspex

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    Yes. And for the horizontal direction?
     
  9. Jan 19, 2013 #8
    ok heres my plan of attack im about to try.

    im going to see if i can calculate how long it takes the projectile to travel 12 meters horizontal, and then im going to see if i can figure out how high it will be after that amount of time.
     
  10. Jan 19, 2013 #9

    SammyS

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    Good plan !
     
  11. Jan 19, 2013 #10

    haruspex

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    That's the way.
     
  12. Jan 19, 2013 #11
    hmm this seems to have failed but also seems close.

    i did d=vot +1/2at2 and solved for t,
    t=[2(d-vo)]/a
    t=2(12m-15m/s)/-9.8m/s2
    t=0.612seconds

    then i attempted to get the vertical distance after, using vertical velocity (30sin60):
    d=vot
    d=[(30)(sin60)]*(0.612)

    and got d=15.91 approx

    the answer in the back of the book is 17.7~ meters so im still off. im guessing one of the two equations i used is incorrect. preliminary suspicions aimed at d=vt probably due to gravity effects.
     
    Last edited: Jan 19, 2013
  13. Jan 19, 2013 #12
    hmm i got another answer using d=vot +1/2at2 again but for vertical, and i got d=14.07m
    still off

    im going to try switching the equations around. d=vt for horizontal and d=vt+1/2at^2 for vertical.
     
  14. Jan 19, 2013 #13
    lol ok nope that failed as well, and i got vertical d=23.92m by that method.
     
  15. Jan 19, 2013 #14
    i think im spinning my wheels here
     
  16. Jan 19, 2013 #15
    im gaining confidence that the time to go 12m is 0.8 horizontal:
    d=vt
    12m=15m/s*t
    t=0.8

    so i just need to get the right vertical distance...
    a=-9.8m/s
    d=?
    t=0.8
    v original=30sin60=25.981~
    v final=?
    a,t,vo=d

    so it looks like i really need to use d=vo(t) +1/2at2

    but why does that give 23.9208m and not 17.7 ??
     
  17. Jan 19, 2013 #16
    The 12m the question refers to is the horizontal displacement of the projectile, not the vertical one. You used the equation for vertical motion of the pebble.
     
  18. Jan 19, 2013 #17
    Are you sure you're using the right velocity value in the vertical equation?
     
  19. Jan 19, 2013 #18
    thats right, i was thinking illogically with that attempt.

    i now believe the horizontal component is simply d=v*t which solves t for 0.8seconds to reach 12m

    i dont know what to do with the vertical distance though. i am frustrated now.
     
  20. Jan 19, 2013 #19

    i dont see how that would be incorrect.

    it is for the horizontal component. the projectile is moving horizontally 12 meters with velocity 15m/s because the actual force is 30m/s 30deg to vertical, thus horizontal force is (30cos60) = 15m/s
    therefore it will take 0.8 seconds to go 12meters
     
  21. Jan 19, 2013 #20
    Where'd the 60 come from?
     
    Last edited: Jan 19, 2013
  22. Jan 19, 2013 #21
    the pebble is slingshotted 30 degrees to the vertical at 30meters/sec

    so the resultant force is 30m/s aimed 30 deg to the vertical (aka 60 degrees to the horizontal, aka vertical force = 30sin60=25.981m/s and horizontal force = 30cos60 = 15m/s
    or if you draw the triangle the other way the Y component is 30cos30 and the X is 30sin30
     
  23. Jan 19, 2013 #22
    Sorry, my bad. I meant your velocity here is weird. How'd you get 15m/s for this if 15m/s is the velocity in horizontal direction? Also how is a metre subtracting a metre/second?
     
  24. Jan 19, 2013 #23
    that equation is incorrect because a=-9.8 is not true in the horizontal motion.
     
  25. Jan 19, 2013 #24
    What numbers do you plug in for vertical motion?
     
  26. Jan 19, 2013 #25
    i dont know, i am very confused and frustrated at the moment.

    the max height of the ball is vf2 = vo2 + 2ad
    vf = 0
    vo = 30sin60 or 30cos30 = 25.981m/s
    a=-9.8m/s
    d=unknown this is what we are solving for
    t=unknown

    vf2 = vo2 + 2ad
    d=34.44m

    after this, we want to see how high the ball will be after it travels 12m horizontally. this is where i am confused.

    my current best attempt is distance=velocity*time (d=vot)
    d= 12m
    vo= 30cos60 or 30sin30 = 15m/s
    t= unknown this is what we are solving for
    i am not worrying about any other variables because i dont see them relevent

    d=vot
    t=0.8s

    this is where i get ultra confused because i see the answer in the back of the book is that the pebble impacts the wall at a height of 17.7m but there is no equation i can solve for d=17.7m

    i feel that the answer should be d= vot + 0.5at2
    so i am very confused what the correct method is.
     
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