1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mum trying to help

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Arrow show with initial velocity at a bird 75m away. Misses.At what points does the arrow pass the bird?


    2. Relevant equations



    3. The attempt at a solution

    Sorry,dont have any idea. An explanation would be most helpful
     
  2. jcsd
  3. Oct 21, 2012 #2
    Is that what the problem says, word for word? It seems like there should be more information, as it's not really clear what's being asked for.
     
  4. Oct 21, 2012 #3
    Yes, poorly worded. An arrow is shot with an initial velocity of 42m/sec at a bird 75m away.The arrow misses the bird. At what time points does the arrow pass the bird?
    With my limited knowledge, I thought the arrow wouldn't travel, if gravity is pulling it down at 9.8m/sec
     
  5. Oct 21, 2012 #4
    I don't understand the question. What is meant by "what time points does the arrow pass the bird"?
     
  6. Oct 21, 2012 #5
    I meant after how many seconds?
     
  7. Oct 21, 2012 #6
    Ok, I'll set it up as though they gave you an angle, and then if they in fact didn't, I'll follow up and show you how to work with an unknown angle.

    Known variables, assuming the arrow is shot at an angle [itex]\theta[/itex] above the horizonal:
    Using basic trigonometry:
    [itex] v_{o} = 42 m/s [/itex]
    [itex] v_{x} = 42 cos \theta[/itex]
    [itex] d_{x} = 75m [/itex]

    Solving:
    We know that distance = velocity x time, thus:
    [itex] 75 = t 42 cos \theta[/itex]
    This gives:
    [itex] t = \frac{75}{42 cos \theta}[/itex]

    Now, if they gave you an angle that the arrow was shot at, this is all you need to do. If they didn't, we will have to use the y-velocity and some other equations to find the minimum angle that the arrow would need to be shot at in order to make it 75m.
     
  8. Oct 21, 2012 #7
    Wouldn't you need to factor in gravity? I was looking at it like this: if the arrow was traveling with an initial velocity of 42m/sec and gravity was pulling it down at a rate of 9.8 m sec, I figured it wouldn't be in the air long enough to travel 75 m. I know I am wrong, but don't understand why
     
  9. Oct 21, 2012 #8
    What I didn't mention, since I assumed we didn't need to worry about the angle, was that I broke the velocity into two components, vertical and hortizonal. Gravity only acts on the arrow vertically, so the horizontal velocity doesn't change. Also, gravity doesn't pull the arrow down 9.8 meters every second, it causes the arrow to accelerate downward [itex] 9.8 \frac{m}{s^{2}} [/itex]

    This relationship is given by:

    [itex] v_{y} = v_{y0} - gt [/itex]
    where [itex] g = 9.8 \frac{m}{s^{2}} [/itex]
    and [itex] v_{yo} = initial-y-velocity = v_{intial}sin\theta = 42 sin\theta [/itex]

    then using the distance formula, you can figure out how long a given arrow will be in the air:

    [itex] d_{y} = 42 sin\theta t - \frac{1}{2}gt^{2} [/itex]

    You would want to find when [itex] d_{y} [/itex] = 0, which happens at t = 0 and one other time. That second time is how long the arrow will be in the air.

    If you solve this you find that order to reach the bird at 75m. That time must be [itex] t \geq \frac{75}{42cos\theta} [/itex]

    The other part I was going to show was that by plugging this time back into the distance formula, the minimum angle the arrow could be shot at and still reach the bird is about 12.3° elevation off the ground.
     
  10. Oct 21, 2012 #9
    Thank you so much for taking the time. I don't pretend to follow everything there, but I can walk through it with my son. Thanks again
     
  11. Oct 21, 2012 #10
    No problem at all. If you guys are still confused and it's just a basic intro course, you might just go ahead and assume the angle is 45°, which should make it all look simpler. It's definitely a pretty poorly worded question if they didn't give you that information though.

    And if you're having trouble with visualizing the equations, try drawing a simple picture with arrows for v, v(x), and v(y). They should form a right triangle with v for the hypotenuse. If you draw an arrow for the force of gravity on the arrow, it would point straight down, which should help you see why gravity only affects vertical velocity. Hope that helps!
     
  12. Oct 22, 2012 #11

    PeterO

    User Avatar
    Homework Helper

    At what angle was the arrow fired? It makes a lot of difference.

    Is the bird 75m away on the ground or 75m directly above you or some unknown height above a point on the ground which is 75 metres away from you???

    Also, just because the force of gravity accelerates the arrow down doesn't necessarily mean the arrow will move down.
    If you apply the brakes in a moving car, the car accelerates backwards, but it does not move backwards, It merely moves forwards more slowly that it was previously [that's what braking is all about!!
    Similarly, the arrow will just be "less high" than it would have been if there was no gravity.
     
  13. Oct 22, 2012 #12

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    I got same answer as bosman27..

    Assumptions:

    Bird is at same elevation as archer
    No air resistance
    Arrow only just misses bird

    In the vertical axis..

    Vy = Vyo - gt
    Vy = 42sin(θ) - gt

    At the top of the flight:

    Vy=0
    and
    t = half the total flight time T.

    Therefore

    0 = 42sin(θ) - gT/2

    rearrange to give...

    T = 2 * 42* sine(θ) / g ....................................................(1)

    In the horizontal axis:

    time = distance/velocity

    T = 75/42cos(θ)..............................................................(2)

    Horizontal and vertical flight times is obviously the same T so equate (1) and (2)

    75/42cos(θ) = 2 * 42 * sine(θ) / g

    rearrange

    75g / (2*422) = sin(θ)cos(θ)

    Now sin(θ)cos(θ) = 0.5 sin(2θ)

    so

    75g/(2*422) = 0.5 sin(2θ)

    θ = 0.5 Sin-1 (75g/422)

    θ = 12.32 degrees

    Then put θ = 10.6 back into (2)

    T = 75/42cos(12.32)

    T = 1.87 seconds.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Mum trying to help
  1. Trying to help my son (Replies: 4)

  2. Please help I tried (Replies: 12)

Loading...