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Hit the Monkey

  1. Jan 23, 2005 #1
    Im having issues with this problem...

    A zookeeper with a tranqualizer dart gun and a monkey (1.5kg) are both 25 m above the ground in trees 90 m apart. Just as the zookeeper shoots the gun, the monkey drops from the tree. What must the minimum muzzle velocity of the dart have been for it to hit the monkey before reaching the ground?

    Im not sure how to start this...so the monkey would be falling at 1/2 gt^2. And at sometime the dart and the monkey will be at the same Y. How would i set up the equations to solve this? Thanks
  2. jcsd
  3. Jan 23, 2005 #2
    The higher the muzzle velocity is, the farther the dart will go (horizontally). Which also means that, the higher the muzzle velocity is, the higher the dart will be when it reaches the tree (25 metres away). Keeping that into consideration, where would the dart and monkey be if the dart hit the monkey with the lowest possibly muzzle velocity? Figure that out and you should be well on your way.
  4. Jan 23, 2005 #3
    The monkey and the dart would probably be right above the ground at the point of contact. So...if the tree is 90 meters high, then i can use -90 as the Y in the height function to find t, and then plug that t into x = [v cos(a)]t, does that seem right?
  5. Jan 23, 2005 #4
    You're almost there. Now, you need to find a. The question tells you what it is. Pay close attention to the words used. What would it be?
  6. Jan 23, 2005 #5
    From your book, you should know that the velocity in the x and y are separate.

    You solved the first part with knowing the amount of time it takes for the monkey to fall. All you need now is to know the amount of time it takes for a dart to go 90 meters.
  7. Jan 23, 2005 #6
    What you started out with inquiring was correct:

    [tex]y = \frac{1}{2}gt^2 [/tex]

    I was semi wrong in stating you did not need this equation, you know that the hunter is shooting straight meaning the angle ([tex]a[/tex]) is [tex]0[/tex].

    [tex]x = vt\cos{a} \rightarrow x = vt\cos{0} \rightarrow x = vt[/tex]

    With solving for [tex]t[/tex] you get the following:

    [tex] t = \sqrt{ \frac{2y}{g} }[/tex]

    In knowing that x and y are separate, you know that the velocity in the x direction is the following:

    [tex]x = vt[/tex]

    In combining the two equations you will get

    [tex] v = x\sqrt{ \frac{g}{2y} }[/tex]
    Last edited: Jan 23, 2005
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