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Help Projectile Motion- Firing of a gun

  1. Sep 26, 2008 #1
    1. A projectile is fired from a gun (on the ground)and has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively.a) How long does it take the projectile to reach the highest point on the trajectory? b) What is the magnitude of the projectile's velocity just before striking the ground? c) How far does the projectile travel horizontally before it hits the ground?

    2. Relevant equations

    3. The attempt at a solution a) t= vy-viy/g 0-40/9.8= 4.08s
    b) v= square root of viy^2 + vix^2= 50
    v=viy+gt 0-9.8*4.08=40 m/s
    c) total time = 2*4.08, 30*8.16= 244.8 m
  2. jcsd
  3. Sep 26, 2008 #2


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    Staff Emeritus
    Science Advisor

    You and reverse should get together!

    (a) Yes, at maximum height the y component of velocity is 0 and, at acceleration g, it will take (vy- viy)/g seconds to reach that height. (I don't know why you switched to 0-40 instead of 40-0 obviously the time is NOT negative.)

    (b) You've actually calculated the speed (magnitude of velocity) initially but, yes, because of the symmetry that is the same as the speed at the end of the trajectory.

    (c) I don't know why you wrote "v=viy+gt 0-9.8*4.08=40 m/s". Obviously, 0-9.8*4.08= -40 m/s, not 40 m/s, but in any case, that is the vertical component of speed just as the projectile hits the ground- which you were not asked. Yes, again because of the symmetry of the trajectory, the total time is exactly twice the time to maximum height and the horizontal distance is just the horizontal component of velocity multiplied by that time.
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