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Projectile motion arrow question

  • Thread starter Lalasushi
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  • #1
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projectile motion question...plz help

hi peeps, im reviewing the projectile motion stuff and for some reason im finding this question really tricky so can anyone take me through this question or at least give me a hand?

The archer fish spits down a prey from overhanging foliage. If the inital velocity of the spit is u = 4.6m/s at an angle of 60 degrees to the surface of the water.
1. What is the height of a bug at the top of the trajectory?
2. How long has the bug got to escape the impending spittle?
3. The bug gets knocked off its branch with a horizontal velocity of 1.15m/s.Given that the horizontal dist. from the spitting position to the branch 0.94m, how far must the fish swim from its spoitting position to catch the fish as soon as it hits the water.

thanks
 

Answers and Replies

  • #2
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Well, it first of all helps to understand what the hell this fish is doing. For that the following link should be helpful.... http://www.naturia.per.sg/buloh/verts/archer_fish.htm [Broken]

You then have to note that the question is making the assumption that the bug will be at the highest point of the trajectory of the spit.

Once you have these two bits of information the problem should be very simple. Much liking working out any other problem in which a ball say is thrown in the air.

Hope this will set you on the right trackl,

H_man

:wink:
 
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  • #3
HallsofIvy
Science Advisor
Homework Helper
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You should know the equations for the horizontal and vertical components of position for a given acceleration (here vertical acc is -9.18 m/s2 and horizontal is 0), and given initial velocity (here the vertical component is 4.5cos(60) and the horizontal component is 4.5 sin(60)). The "vertical" equation is quadratic and its maximum can be found by completing the square.
 

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