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Monkey Shoot

  1. Jan 7, 2008 #1
    [SOLVED] Monkey Shoot

    1. you aim in a direct line at a monkey hanging on a tree branch with a tranq gun. As soon as you fire, that instant, the monkey lets go. Do you hit the monkey? Prove your calculations through algebra and knowledge of physics



    2. Money: D1-D0 = -1/2gt^2
    Rifle: D1-D0 = (V0*sintetha)t-1/2*g*t^2




    3. ok, I know that the projectile hits the monkey. Because the only force acting on the monkey and the projectile is the force of gravity. So, they both will have the same vertical distance drop. And thus, the projectile and monkey will be at the same location and cause the monkey to get hit. I managed to figure it out by physics, but can someone please explain this to me in an algebra way. Thanks.
     
  2. jcsd
  3. Jan 7, 2008 #2
    When you say the projectile hits the monkey, you're implying that the projectile and the mokey's vertical displacement is the same after a time t.

    Your 2nd equation is not correct because it looks like you're modeling the vertical distance, but you included Vo which is a horizontal component (because you shoot at a straight horizontal line).
     
  4. Jan 7, 2008 #3

    Shooting Star

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    Same at the time t if the dart takes time t to cross the initial horizontal distance.
     
    Last edited: Jan 8, 2008
  5. Jan 7, 2008 #4
    Oh and we have been given the varibles that range is D and height is H.

    and I was given that formula...Vertical projectile Y-Distance is (V0SinTheath)t-1/2*g*t^2

    so i need an algebra explanation with varibles(D-range,H-height the monkey starts out at, to prove this...Why the projectile stricks the monkey.

    Can someone help please?

    Here is a bad picture of what im talking about:
    [​IMG]
     
    Last edited: Jan 7, 2008
  6. Jan 7, 2008 #5

    hage567

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    You need to find the time it takes the bullet to travel the horizontal range D. Do you know how to state the horizontal component of the bullet's initial velocity?
     
  7. Jan 7, 2008 #6
    ok horizontal velocity its V0cos(theta)
    &
    the formula is X-distance = (V0costetha)t

    so t = d/(V0costetha)

    Im getting these equations, i solved for time and other stuff, I pluged them into each other and everything... I was just running around in circles... at one point I got H = (V0sintetha)t. But, I don't know what it means.

    And how is velocity related to this?
     
  8. Jan 7, 2008 #7
    you aim in direct line, so your theta should be zero. Plug it in your equation in post #4.
     
  9. Jan 8, 2008 #8

    Mute

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    The rifle is aimed in a direct line at the monkey. Theta is not zero.
     
  10. Jan 8, 2008 #9
    but the monkey is elevated...a height of H. So how is the theta 0, can someone explain please?
     
  11. Jan 8, 2008 #10

    Shooting Star

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    Theta is not zero, as has been pointed out in post #8.
     
  12. Jan 8, 2008 #11

    Shooting Star

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    tan(theta) = H/D.

    If the projectile (tranq dart) takes time t to traverse a horizontal distance of D, then the eqns of motion of the dart are:

    D = vcos(theta)*t,
    y = vsin(theta)*t – (1/2)gt^2. So,

    y = vsin(theta)*D/[vcos(theta)] - (1/2)gt^2 =>
    y = H - (1/2)gt^2. (Since, H/D = tan(theta).)

    This shows that at time t, the dart has gone a horizontal distance of D and a vertical distance of H - (1/2)gt^2 from the ground.

    But in time t, the monkey has fallen (1/2)gt^2, and its position is H - (1/2)gt^2 from the ground.

    So, at time t, the dart and the monkey are at the same point.
     
  13. Jan 8, 2008 #12
    Thanks for the explaination, but I am a little confused on how vsin(theta)*D/[vcos(theta)] = H? Because you got H - (1/2)gt^2 for the vertical height for both the monkey and the projectile.

    EDIT: THANKS I got it!!! I just looked at it wrong. THANKYOU THANKYOU. I spent 2 days and couldn't prove it! thanks
     
    Last edited: Jan 8, 2008
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