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Please help on one mroe physics concept

  1. Jun 25, 2006 #1
    A child kicks a ball with an initial velocity of 8.5 meters per second at an angle of 35° with the horizontal, as shown. The ball has an initial vertical velocity of 4.9 meters per second and a total time of flight of 1.0 second. [Neglect air resistance.]
    what is the max height? what is the horizontal componenet of the ball's initial velocity?

    A projectile is fired from a gun near the surface of Earth. The initial velocity of the projectile has a vertical component of 98 meters per second and a horizontal component of 49 meters per second. How long will it take the projectile to reach the highest point in its path?

    please show formula and answer.

    thanks, and these are my last questions bw.
     
  2. jcsd
  3. Jun 25, 2006 #2
    You're going to use the same equations as the last question you asked.

    Remember:

    V(t) = At + Vo
    X(t) = (1/2)At^2 + Vot + Xo

    Vo = Initial Velocity
    Xo = Initial Position

    These are good for any axis.

    You should try going through the problem real quick and looking at how your past questions were solved.
     
  4. Jun 25, 2006 #3
    ok for the first problem was ur max height 4.9 m.

    and for problem two did u get 7 m/s as the horizontal component of the initial point? u dont have to tell me the answer if its wrong, jus tell me if its wrong cuz i think i put a negaive where i shouldnt have
     
  5. Jun 25, 2006 #4
    I think something is wrong with your first answer. How are you setting up your formulas?

    I got 7 m/s for the horizontal component though.
     
  6. Jun 25, 2006 #5
    nvm i got 8.5 for the first one, but if it isnt that then it is 1.2 it depends which i was supposed to use for the firs component which im still confused about
     
  7. Jun 25, 2006 #6
    One of those answers is what I got. Show me how you're setting up your formulas.
     
  8. Jun 25, 2006 #7
    my Voy=4.9, my V=8.5

    so i did -5.73=-12.1+9.8t, 6.37=9.8t, =0.65 SO

    then i used the 3rd equation so yea its 1.2m
     
  9. Jun 25, 2006 #8
    I'm a little confused about how you're using the formulas but you came out with the right answer. I'll show you how I did it.

    Our formula for height is going to be:

    Y(t) = -5t^2 + 4.9t + 0

    I rounded gravity (9.8) to 10 and 1/2 of that is 5. I usually deal with gravity as being a negative force so that's why the negative is out in front (A = -10 m/s^2), 4.9 just comes from our Vo (initial velocity).

    Now we know total flight time is 1 second and we know that it takes the same amount of time to go up as it does to come down so the max height will occur at .5 seconds. Plug that in for t and you've got 1.2 m.
     
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