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*please Check My Asnwer! :'(

  1. Sep 26, 2004 #1
    *******please Check My Asnwer! :'(********

    Q40. A boy stands at the edge of a bridge 20.0 m above a river and throws a stone straight down with a speed of 12.0 m/s. He throws another pebble straight upward with the same speed so that it misses the edge of the bridge on the way back down and falls into the river. For each stone find (a) the velocity as it reaches the water and (b) the average velocity while it is in flight.Note: Ignore the affects of air resistance.

    (a) X= 20 m a=9.80m/s2 Vi=12.0 m/s Vf= ????
    V_f^2 = V_i^2 + 2a (ΔX)
    V_f^2 = (12.0 m/s)^2 + 2(9.80 m/s^2) (20 m)
    V_f^2 = (144 m^2/s^2) + (19.6 m/s^2) (20 m)
    V_f^2 = (144 m^2/s^2) + (392 m^2/s^2)
    V_f^2 = (536 m^2/s^2)
    V_f = 23.15167381 m/s
    V_f = 23 m/s




    Therefore the velocity of the stone as it reaches the water is approximately 23 m/s. Since the pebble was thrown straight upward with the same speed so that it misses the edge of the bridge on the way back down and falls into the river, I know that if the pebble is projected upwards in a vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity it has when it returns to the same height. That is, a pebble projected vertically with an upward velocity of +12 m/s will have a downward velocity of –12 m/s when it returns to that same height. So the velocity of the pebble as it reaches the water is –12 m/s.
     
  2. jcsd
  3. Sep 26, 2004 #2
    I Dont Understand How To Do Part B Can Someone Help Me!??
     
  4. Sep 26, 2004 #3

    enigma

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    I'm pretty sure this is a typo, and everything else looks good.

    For part (b), you need to divide the net distance travelled (the 20m fall) by the total time the trip took. This will obviously be different for the two stones.
     
  5. Sep 26, 2004 #4
    So the velocity of the pebble as it reaches the water is –12 m/s.
    thats wrong?????? :{
     
  6. Sep 26, 2004 #5
    how do i find the time for the stone and pebble tho?
     
  7. Sep 26, 2004 #6
    which kinematic equation shud i use to calculate time for the stone n pebble, theres four so do i just choose one and rearrange the equation to solve for t and substitute the valuesin ????? help
     
  8. Sep 26, 2004 #7
    for part b the asnwer id -12 m/s right negative cuz of the direction?
     
  9. Sep 26, 2004 #8
    The (time) averaged value of the velocity (denoted [tex]<v(t)>[/tex])is given by

    [tex]<v(t)> = \frac{1}{T} \int_{0}^{T} v(t) dt [/tex]

    Here, [tex]T[/tex] is the total time of flight. In this case, the time from when the stone leaves the boy's hand to right before it hits the water. For both cases (stone thrown up and stone thrown down) you will have a different [tex]T[/tex]. Find this value for both! Then use [tex]v(t) = v_{0} - gt[/tex] for the integral.

    Or if you want to save a lot of time you can just use

    [tex]<v> = \frac{1}{2}(v_{f} + v_{0})[/tex]

    (it's the same thing as doing the integral). Make sure you use the correct signs here!
     
  10. Sep 26, 2004 #9
    i dun know what that integral thing isi never learned that?
     
  11. Sep 26, 2004 #10

    enigma

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    If you defined 'up' as positive, then you will have a negative average velocity, yes.

    If it's got a velocity of 12m/s on the way back down, the same as if you threw it straight down, how would they have different velocities at the bottom?
     
    Last edited: Sep 26, 2004
  12. Sep 26, 2004 #11
    Q40. A boy stands at the edge of a bridge 20.0 m above a river and throws a stone
    straight down with a speed of 12.0 m/s. He throws another pebble straight
    upward with the same speed so that it misses the edge of the bridge on the way
    back down and falls into the river. For each stone find (a) the velocity as it
    reaches the water and (b) the average velocity while it is in flight.
    Note: Ignore the affects of air resistance.

    (a) X= 20 m a=9.80m/s2 Vi=12.0 m/s Vf= ?????
    Vf2 = Vi2 + 2a (ΔX)
    Vf2 = (12.0 m/s)2 + 2(9.80 m/s2) (20 m)
    Vf2 = (144 m2/s2) + (19.6 m/s2) (20 m)
    Vf2 = (144 m2/s2) + (392 m2/s2)
    Vf2 = (536 m2/s2)
    Vf = 23.15167381 m/s
    Vf = 23 m/s

    Therefore the velocity of the stone as it reaches the water is approximately 23 m/s. Since the pebble was thrown straight upward with the same speed so that it misses the edge of the bridge on the way back down and falls into the river, I know that if the pebble is projected upwards in a vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity it has when it returns to the same height. That is, a pebble projected vertically with an upward velocity of +12 m/s will have a downward velocity of –12 m/s when it returns to that same height.

    (b) *AVERAGE VELOCITY*= ΔX/Δt

    STONE
    Xf = 20 m Vxf = 23 m/s Vxi=12.0 m/s ax=9.80m/s2 Xi = 0 m t=????

    Vxf = Vxi + ax t Vavg= ΔX/Δt
    23 m/s = 12.0 m/s + (9.80 m/s2) t Vavg= 20m/1.12244898 s
    23 m/s – 12.o m/s = (9.80 m/s2) t Vavg= 17.81818181 m/s
    11 m/s = (9.80 m/s2) t Vavg= 17.81 m/s
    t = 11 m/s .
    9.80 m/s2

    t = 1.12244898 s
    t = 1.12 s
     
  13. Sep 26, 2004 #12
    am i on the right track?
     
  14. Sep 26, 2004 #13
    OKAY I TRIED IT AGAIN
    Q40. A boy stands at the edge of a bridge 20.0 m above a river and throws a stone
    straight down with a speed of 12.0 m/s. He throws another pebble straight
    upward with the same speed so that it misses the edge of the bridge on the way
    back down and falls into the river. For each stone find (a) the velocity as it
    reaches the water and (b) the average velocity while it is in flight.
    Note: Ignore the affects of air resistance.

    (a) X= 20 m a=9.80m/s2 Vi=12.0 m/s Vf= ?????
    Vf2 = Vi2 + 2a (ΔX)
    Vf2 = (12.0 m/s)2 + 2(9.80 m/s2) (20 m)
    Vf2 = (144 m2/s2) + (19.6 m/s2) (20 m)
    Vf2 = (144 m2/s2) + (392 m2/s2)
    Vf2 = (536 m2/s2)
    Vf = 23.15167381 m/s
    Vf = 23 m/s

    Therefore the velocity of the stone as it reaches the water is approximately 23 m/s. Since the pebble was thrown straight upward with the same speed so that it misses the edge of the bridge on the way back down and falls into the river, I know that if the pebble is projected upwards in a vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity it has when it returns to the same height. That is, a pebble projected vertically with an upward velocity of +12 m/s will have a downward velocity of –12 m/s when it returns to that same height.

    (b) *AVERAGE VELOCITY*= ΔX/Δt

    STONE
    Xf = 20 m Vxf = 23 m/s Vxi=12.0 m/s ax=9.80m/s2 Xi = 0 m t=????

    Vxf = Vxi + ax t Vavg= ΔX/Δt
    23 m/s = 12.0 m/s + (9.80 m/s2) t Vavg= 20m/1.12244898 s
    23 m/s – 12.o m/s = (9.80 m/s2) t Vavg= 17.81818181 m/s
    11 m/s = (9.80 m/s2) t Vavg= 17.81 m/s
    t = 11 m/s .
    9.80 m/s2

    t = 1.12244898 s
    t = 1.12 s



    (b) *AVERAGE VELOCITY*= ΔX/Δt

    PEBBLE
    Xf = 20 m Vxf = 12.0 m/s Vxi=12.0 m/s ax=9.80m/s2 Xi = 0 m t=????
    Vxf = Vxi + ax t
    12.0 m/s = 12.0 m/s + (9.80 m/s2) t
    12.0 m/s – 12.0 m/s= (9.80 m/s2) t
    0 m/s = (9.80 m/s2) t
    0 m/s = t
    9.80 m/s2
    t = 0 m/s
     
  15. Sep 26, 2004 #14

    enigma

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    It looks like you're getting there. You're still a little confused with your positive/negative velocity convention.

    Draw your picture, and define a direction as 'positive', whether it's up or down. Now make sure that all of the negative signs work out.

    If you set 'up' as positive and the cliff as the origin, then the acceleration of gravity is negative, the displacement (20m) is negative, the initial velocity of the thrown downward ball is negative, and the initial velocity of the thrown upwards ball is positive.

    Now: What was the ending velocity of the thrown upwards ball? You pulled your conclusion out without correcting it. Is there any difference between the upwards ball when it's coming back down and passing the origin and the downwards thrown ball? Will they have the same or different velocities at the bottom?

    What you've got shown now is that the ball you throw upwards decelerates, turns around, comes back to the cliff level, and then keeps going at the same speed downward with no gravity acting on it.

    PS: The forum code parses multiple spaces in a single line, so it's a bit hard to read your calculations. Either put each calculation on a seperate line, or use the {code} {/code} tags, replacing {} with []
     
    Last edited: Sep 26, 2004
  16. Sep 26, 2004 #15
    (b) *AVERAGE VELOCITY*= ΔX/Δt

    PEBBLE
    Xf = 20 m Vxf = 12.0 m/s Vxi=12.0 m/s ax=9.80m/s2 Xi = 0 m t=????

    Vxf = Vxi + ax t Vavg= ΔX/Δt
    12.0 m/s = 12.0 m/s + (9.80 m/s2) t Vavg= 20 m/ o s
    12.0 m/s – 12.0 m/s= (9.80 m/s2) t Vavg= LOL???
    0 m/s = (9.80 m/s2) t
    0 m/s = t
    9.80 m/s2
    t = 0 m/s
     
  17. Sep 26, 2004 #16

    enigma

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    You need to be careful here. The units of time are not m/s.

    Units follow the rules of algebra also. It's the easiest way to find out if you've done a mistake.

    [tex]\frac{m}{s}=\frac{m}{s^2} \times s[/tex]

    [tex]\frac{m \times s^2}{s \times m}= s[/tex]

    [tex] s = s[/tex]
     
  18. Sep 26, 2004 #17
    sorry it is +12 m/s my mistake
     
  19. Sep 26, 2004 #18
    PEBBLE
    Xf = 20 m Vxf = 12.0 m/s Vxi=12.0 m/s ax=9.80m/s2 Xi = 0 m t=????

    Vxf = Vxi + ax t Vavg= ΔX/Δt
    12.0 m/s = 12.0 m/s + (9.80 m/s2) t Vavg= 20 m/ o s
    12.0 m/s – 12.0 m/s= (9.80 m/s2) t Vavg= LOL???
    0 m/s = (9.80 m/s2) t
    0 m/s = t
    9.80 m/s2
    t = 0 s
     
  20. Sep 26, 2004 #19
    :uhh: yeah sorry mistake it is seconds thanks for the tip, i forgot the units cancel out
     
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