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1018

  • one year ago

The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 19 m/s. (a) What is the magnitude of the velocity of the projectile 1.7 s before it achieves its maximum height? (b) What is the magnitude of the velocity of the projectile 1.7 s after it achieves its maximum height? If we take x = 0 and y = 0 to be at the point of maximum height and positive x to be in the direction of the velocity there, what are the (c) x coordinate and (d) y coordinate of the projectile 1.7 s before it reaches its maximum height and the (e) x coordinate and (f) y coordinate

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  1. IrishBoy123
    • one year ago
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    the first sentence tells you the projectile's horizontal velocity, which you will assume to be constant in this problem and that point you also know that its vertical velocity = 0 and that, more generally, acceleration in the vertical direction is -g. armed with that information, all you need are the equations of motion to solve https://gyazo.com/002724abd9cb7eb055122e07be79d344

  2. 1018
    • one year ago
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    thanks! im going to try it out

  3. 1018
    • one year ago
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    @Michele_Laino this one

  4. 1018
    • one year ago
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    i hate to ask, but can you also solve it? its worth alot of points on my homework and i hate to waste it

  5. Michele_Laino
    • one year ago
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    ok!

  6. 1018
    • one year ago
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    thank you so much!

  7. Michele_Laino
    • one year ago
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    here we have a parabolic motion, the situation is like below: |dw:1440003729324:dw|

  8. 1018
    • one year ago
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    oh wait! sorry!

  9. 1018
    • one year ago
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    The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 14 m/s. (a) What is the magnitude of the velocity of the projectile 1.2 s before it achieves its maximum height? (b) What is the magnitude of the velocity of the projectile 1.2 s after it achieves its maximum height? If we take x = 0 and y = 0 to be at the point of maximum height and positive x to be in the direction of the velocity there, what are the (c) x coordinate and (d) y coordinate of the projectile 1.2 s before it reaches its maximum height and the (e) x coordinate and (f) y coordinate 1.2 s after it reaches its maximum height?

  10. Michele_Laino
    • one year ago
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    at that point the kinetic energy is: \[\Large KE = \frac{1}{2}m{v^2}\] where v is the velocity in my drawing, which is a horizontal vector

  11. 1018
    • one year ago
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    im sorry but it's this one. its actually the same but with different values sorry didnt see that

  12. Michele_Laino
    • one year ago
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    ok! I see

  13. Michele_Laino
    • one year ago
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    we have to compute the velocity when our projectile, passes at these points: |dw:1440004078499:dw|

  14. Michele_Laino
    • one year ago
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    namely at P1, and P2

  15. 1018
    • one year ago
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    ok. im following so far

  16. Michele_Laino
    • one year ago
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    here we have to keep in mind that the initial vertical velocity Vy is given by the subsequent formula: \[\Large \frac{1}{2}mv_y^2 = mgh\] where h is the maximum height: |dw:1440004390763:dw|

  17. Michele_Laino
    • one year ago
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    |dw:1440004431440:dw|

  18. 1018
    • one year ago
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    what is m? and is g the gravitational acceleration?

  19. Michele_Laino
    • one year ago
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    m is the mass of our projectile, and g is gravity In a parabolic motion the horizontal velocity, doesn't change, so along all motion, the horizontal velocity is: \[\Large {v_x} = 14m/\sec \]

  20. 1018
    • one year ago
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    yes, and that's all i got so far. at least it's correct. haha

  21. 1018
    • one year ago
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    oh, i'll just see what you do. sorry im not familiar with using your formula, with the m one

  22. Michele_Laino
    • one year ago
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    the equation for the vertical component of the velocity is: \[\Large {v_y}\left( t \right) = {v_{0y}} - gt\]

  23. Michele_Laino
    • one year ago
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    where g is gravity and V0y is: \[\Large {v_{0y}} = \sqrt {2gh} \] which comes from my formula above: \[\Large \frac{1}{2}mv_y^2 = mgh\]

  24. Michele_Laino
    • one year ago
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    so, after a substitution, we get: \[\Large {v_y}\left( t \right) = {v_{0y}} - gt = \sqrt {2gh} - gt\]

  25. Michele_Laino
    • one year ago
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    nevertheless we don't know the value of the maximum height h

  26. 1018
    • one year ago
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    yes

  27. Michele_Laino
    • one year ago
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    I think that we have insufficient data

  28. Michele_Laino
    • one year ago
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    we need to know, at least, the measure of the maximum height

  29. 1018
    • one year ago
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    sorry, but that is everything

  30. Michele_Laino
    • one year ago
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    I think that many parabolic motions are possible with the horizontal velocity which you have provided

  31. 1018
    • one year ago
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    someone answered this a while ago but i saw it late, maybe you can check out what he/she said

  32. Michele_Laino
    • one year ago
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    I saw that answer

  33. Michele_Laino
    • one year ago
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    I think that with your data, the motion it is not uniquely determined

  34. 1018
    • one year ago
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    any other ideas how to do this?

  35. Michele_Laino
    • one year ago
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    I'm sorry, no my idea is that we need to know the maximum height or another parameter, for example the angle between the initial velocity and the horizontal line |dw:1440006104583:dw|

  36. 1018
    • one year ago
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    that's ok. thank you so much anyway. hey if you're interested, i could printscreen the solution. it shows the solution when after you answered incorrectly. haha

  37. 1018
    • one year ago
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    im gonna move on from this, try to get some rest, study again tomorrow. haha

  38. Michele_Laino
    • one year ago
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    yes! I'm interested, please tell me your solution :)

  39. 1018
    • one year ago
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    ok, wait. i'll just make the answer wrong haha

  40. Michele_Laino
    • one year ago
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    ok!

  41. 1018
    • one year ago
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    here

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  42. 1018
    • one year ago
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    it's a printscreen of the solution

  43. Michele_Laino
    • one year ago
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    yes! that is the solution, and I have failed, instead. Sorry!!

  44. 1018
    • one year ago
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    no no its ok. it's not like i figured it out too. haha. thank you so much anyway

  45. Michele_Laino
    • one year ago
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    ok! :)

  46. 1018
    • one year ago
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    hey, im gonna rest for awhile, i'll review this again tomorrow. haha. thank you again!

  47. Michele_Laino
    • one year ago
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    :)

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