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Resultant Velocity of a Projectile

  1. Jul 19, 2013 #1
    1. The problem statement, all variables and given/known data

    An archer stationed at the edge of a cliff shoots an arrow horizontally at 90ms^-1. The arrow falls down and hits a target. Assuming that the arrow is shot from a height of 20m, calculate:

    i. How long it takes the arrow to reach the target. (done)
    ii. How far horizontally the target is from the base of the cliff. (done)
    iii. The resultant velocity at impact and the angle this makes with the vertical. Mention one assumption made.

    2. Relevant equations

    SUVAT equations:


    3. The attempt at a solution

    Parts i. and ii. were basically tackled by splitting the problem in half via resolving components vertically and horizontally. Time taken for the arrow was found to be 2s and the target was found to be 180m away from the base of the cliff.

    What confuses me in part iii. is the assumption I have to make. I know that resultant velocity is the resultant vector from the horizontal and vertical components, and that the angle is found by resolving the vector diagram.

    Am I missing something obvious? Cheers.
  2. jcsd
  3. Jul 19, 2013 #2
    You're doing it right bro..
    Maybe by the 'assumptions' they mean some basic assumptions like neglecting resistance by air, uniform gravity, etc.
  4. Jul 19, 2013 #3
    Yeah, negligible air resistance is an assumption. (my physics course doesn't feature air resistance at all) But how is this going to help me in finding the resultant velocity at impact?
  5. Jul 19, 2013 #4


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    Homework Helper

    Well there's no horizontal acceleration, so you can take the final horizontal velocity to be 90 m/s.

    You can take the final vertical velocity to be ##\vec{a} Δt##.

    The resultant velocity would then be : ##\vec{v}_R = \sqrt{ (v_{F_V})^2 +(v_{F_H})^2 }##

    Then to get the direction, make your triangle.
  6. Jul 20, 2013 #5
    I think I get it now. The only acceleration present is the downward acceleration due to gravity so obviously the horizontal acceleration would be 0.

    Since impact occurs when t=2s, I made a triangle with 90m/s as the horizontal component and 19.6m/s as the vertical component. This is since after 1s, it's accelerating at 9.81m/s per second and thus after 2s, it becomes 19.6m/s (per second). Resultant Velocity was then worked out to be 92.1m/s and the angle that it makes with the vertical was found out to be 77.7°.

    This is better summed up in this diagram (it has different numbers though, but still relevant)


    Does any of the above make sense? Lastly, cheers for helping out guys :)
    Last edited: Jul 20, 2013
  7. Jul 20, 2013 #6
    There seems a small ambiguity in that sentence.
    Can you find the resultant velocity and angle with vertical after 1.3 sec?
    Just to check if you really got it! :smile:
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