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What is the maximum height achieved by the projectile?

  1. Oct 12, 2004 #1
    I have a question from my practice test that is supposed to be on our midterm, so I want to make sure I am doing it right!

    A 10kg projectile is launched from the wall of a canyon at an angle of 30 degrees above the horizon with a velocity of 100m/s. The wall is 1500m above the canyon floor.
    a)what is the maximum height achieved by the projectile?
    b)what is the impact velocity with the canyon floor?

    So, I did part a and I found the x and y components of v...vsin30 and vcos30.
    I then found the time and used it it the equation y=Voyt + 1/2at^2 and I got y=128m

    So, for part b which equation do I use? If I use pythagorean theorem the answer is 100m/s which is the same as the initial. Is that right? Or do I need to use a different equation?
  2. jcsd
  3. Oct 12, 2004 #2


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    why don't you use conservation of mechanical energy?

    [tex] \Delta K + \Delta \Omega = 0 [/tex]

    For max height, you know Vy = 0, and Vx is constant, so use that. There is gravitational potential energy. You could do it with kinematics, but why bother?
  4. Oct 12, 2004 #3
    Because, we have not learned that yet so I can't use the equation
  5. Oct 12, 2004 #4
    Hmmm, big no on that.

    You split the velocity into x-y components which is good, but your choice of formular for part 1 is somewhat questionable. See the question have not given you how long the bullet stays in the air, and therefore you would have 2 unknowns in d=ut + 1/2 at^2. it's still doable but take much more effort in figuring out the t.

    Since you're only asked to figure out the height, just consider everyting in y direction. Try instead using v^2 = u^2 +2ad. (v==>final velocity, u ==> initial velocity, a ==> acceleration due to gravity, negative because it's trying to stop the bullet, d ==> distance) set v = 0 since the highest point is when the bullet is no longer going up. and give it a try. (you should get some 500ish answer)

    for part b, use the same equation, but set u = 0 since you're calculating the velocity of bullet going from the top of the arc to the ground in the y direction. set d = the answer you have from part a + height of the wall, a = acceleration due to gravity, postive this time because it's accelerating the bullet.

    after you figure out the final velocity in y direction, you can just add it with the x-direction velocity (which haven't changed) using pythagorean theorem.

    Hope that helps
  6. Oct 12, 2004 #5


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    Well in that case, Remember Vy = 0 at max height. if you found the time at max height, then plug it back, show me how you found the time at max height, so i can see if its ok. By the way you got an initial position, Yo = 1500 m
  7. Oct 12, 2004 #6
    Ok so I figured out the height w/o using the time,
    V^2 = Vi^2 +2ay and I still get 128m
    Then I used the same equation to get the final velocity of y and I set the initial velocity to 0 and I got 179m/s,

    Then I used the pythagorean theorem to get the final velocity 199m/s

    Does that seem right?
  8. Oct 12, 2004 #7


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    You're not taking into account the initial Y.
  9. Oct 12, 2004 #8
    I used the same Vx because it stays constant right? Vcos30= 87m/s
  10. Oct 12, 2004 #9


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    Vx stays constant, yes
  11. Oct 12, 2004 #10
    So, was my answer right or am I still doing something wrong inthe final equation?
  12. Oct 12, 2004 #11
    yeah, that's about right, I forgot the Sin whe I did my sketch.
  13. Oct 12, 2004 #12
    I did take into account the initial y... I added it to the y I found in part a for the equation to find the final velocity of y
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