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Does the formula v=∆x/(cos⁡θt) give the TOTAL velocity of a projectile

  1. Jan 27, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm doing an experiment where I'm trying to determine the velocity of a hose when fully turned on using projectile motion. I projected the hose at different angles (15,30,45,60,75,90 degrees) doing 3 tests on each 1. Here is an example I made with the a set of results -

    At an angle of 15 degrees from the ground, a hose is turned on. It has a time of flight of 0.4 seconds and a horizontal range of 2.47 meters.


    2. Relevant equations
    v=∆x/(cos⁡θt)


    3. The attempt at a solution
    Horizontal range = v * cos θ * t
    2.47 = v * cos 15 * 0.4
    v = 2.47 ÷ (cos 15 * 0.4) = 6.392830414 m/s

    Is this the TOTAL velocity of the water coming from the hose?

    I've asked this question before but somebody told me that this was only the initial velocity, not the actual velocity of the water from the hose and now I'm confused.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 27, 2014 #2

    lightgrav

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    where-ever you got that equation, it should not have been presented in that format.
    if you launch an object with velocity v , that is elevated by angle ⁡θ above horizontal,
    then the horizontal displacement during time t is ∆x = (v cos⁡θ) t
    ... the (v cos⁡θ) is just the horizontal velocity component.

    The key issue with projectiles is figuring out how long (t) the thing is in the air.
    on level ground, your 2.47m/s water only stays in the air 0.13s.
     
  4. Jan 27, 2014 #3
    But it wasn't 2.47 m/s, it was 2.47 m. So is the answer still wrong?
     
  5. Jan 27, 2014 #4

    lightgrav

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    yeah, you pretended it spent .4s in the air. ... what do you think its vertical speed was, at the beginning?
     
  6. Jan 27, 2014 #5
    Why do you think I pretended?
     
  7. Jan 27, 2014 #6

    lightgrav

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    sorry, your post #3 did not all display when I posted my #4.

    if the (full) initial speed had been 6.39 m/s, its upward portion would have started at 1.65m/s,
    and it would've spent .34 s in the air , is that close enough to 0.4 s ?

    is this 0.4s measured in the experiment? That would be unusual, and tricky to accomplish with precision.
    (measuring the angle is easy, measuring range with water maybe not)
     
  8. Jan 27, 2014 #7
    I did 3 tests for all of the angles... To time the water I used a camera, filmed it and then slowed it down on my computer.
    For 15 degrees, I got 2.47 m, 2.40 m, and 2.43 m (range of each test, average of 2.43 m) and 0.40 s, 0.38 s, and 0.36 s (time of flight for each test, average of 0.37 s). The average velocity for 15 degrees was 6.64 m/s. I'm just unsure if this is the velocity of the water.
     
  9. Jan 27, 2014 #8

    lightgrav

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    trusting your time and horizontal distance measurements, that equation will provide the entire (diagonal) speed.

    how high above the landing-spot was the nozzle opening? (that will slightly increase the flight time).
     
  10. Jan 27, 2014 #9
    its was about 7 cm above, but I could just include that in my report as error or measure the range from where the nozzle is closest to the ground and include it in the displacement. I know that would be inaccurate but there is room for inaccuracies for this assignment.
     
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