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Homework Help: Projectile being launched from ground level

  1. Sep 25, 2012 #1
    1. The problem statement, all variables and given/known data

    A projectile is fired from ground level with an initial velocity of 50 m/s and an initial angle of 37°. Assuming g = 9.8 m/s^2, find:

    a) the projectiles total time of flight.
    b) the maximum height attained
    c) the total horizontal distance traveled
    d) the final horizontal and vertical velocities just before it hit the ground.

    2. Relevant equations

    okay, so beginning with a), I believe the necessary equation is h=vi2Sin2θi/2g

    3. The attempt at a solution

    Plug in a) into my calculator, and I get uhh 4436.721196. I'm not sure if i'm using the correct method.
  2. jcsd
  3. Sep 25, 2012 #2
    You need to find the total time of flight in a), the equation you posted is for the max. height attained by the projectile.
  4. Sep 25, 2012 #3
    would you utilize Vf=Vi+at?

    So Vf = 0 (at the top of the trajectory Vf would be zero.
    g = -9.8
    Vi = 50

    Plus in so you get -9.8t = -50
    t = 5.102?
  5. Sep 25, 2012 #4
    For the vertical motion, what's the initial velocity? You need to take the component of initial velocity in the vertical direction for Vi.
  6. Sep 25, 2012 #5
    Hmm I got 50*Sin(37) which gave me 30.09075116.
  7. Sep 25, 2012 #6
    You shouldn't get such a big answer. The component you took is correct. Check your calculations. Maybe, it will help if you show us your steps. Make sure you double the time you get when you plug the values in the equation Vf=Vi+at because you need to find the total time of flight. The projectile reaches the max. height and comes back again.
  8. Sep 25, 2012 #7
    Okay, well in my textbook, it says the following formula is utilized to find Vyi:


    I plugged in the numerical values into my calculator and got Vyi = 30.09075116.

    Now, I plugged that answer into Vyf = Vyi + ayt

    Vyf should equal zero because at the top of the trajectory the y component is zero.

    I solved and got t = 3.09075116 seconds.

    Edit; multiply the answer by 2 and you get 6.14096.... Look good?
  9. Sep 25, 2012 #8
    Oops, sorry, i thought you meant time of flight when you wrote 30.09075116. :redface:

    Just double the time you calculated, for the reason i mentioned in my previous post. :wink:

    EDIT: Just saw your edit, looks good.
  10. Sep 25, 2012 #9
    Thanks :)

    If you could, could you just quickly check my work for the rest of them?

    In regards to b), I used the equation I mentioned initially (Vi2Sin2θi/2g

    Plugged in (50^2)*Sin(37)*Sin(37)/2*9.8 and got 4436.721196.

    Look a little off?
  11. Sep 25, 2012 #10
    Its too much off. :biggrin:
    Check your calculations again. Or show the steps.
  12. Sep 25, 2012 #11
    Ah, yeah, incorrect calculations lol. I fixed it and got something around 46.196..

    Okay, so part c) I found the Vxi to be 39.931775 or does this have something to do with range (R)?
  13. Sep 25, 2012 #12
    Yes, you need to find the range.
  14. Sep 25, 2012 #13
    c) Vxi = 39.9317755

    Formula for R being: Vxi2ta

    Ta being the halfway point of the trajectory so all in all this comes out to be the time we calculated earlier.

    R = (39.9317755)(3.07048412)2

    R = 245.2198204 meters

    Look good?
  15. Sep 25, 2012 #14
    Looks correct to me.
  16. Sep 25, 2012 #15
    Okay, lastly, d).

    Hmm okay, well horizontal velocity is constant so I don't think there is a way of doing the horizontal one.

    Wouldn't these just be their respective initial velocities? Or maybe you'd plug them into an equation that has a Vf and solve.

    Vf = Vi + at

    Time being how long the ball is in the air.
  17. Sep 25, 2012 #16
    Horizontal velocity will be the same, vertical velocity will be opposite in direction to that of initial. It asks about velocity, not speed.
  18. Sep 25, 2012 #17
    Understood, so the object fired has an initial velocity of +50 m/s; however, as it comes down, the velocity is the same, but simply decreasing?

    So the velocity just as the object hits the ground is -50 m/s?

    Edit: Leaving for class right now. Thanks for taking the time to explain this to me. I really appreciate the help. :)
  19. Sep 25, 2012 #18
    No, i did not mean that. The question asks about the vertical and horizontal final velocities. The direction of final vertical velocity would be the opposite to that of the initial vertical velocity. You cannot say that its -50 m/s.
  20. Sep 27, 2012 #19
    So what would it be? Would it be 50 m/s simply on the opposite direction? In other words, 50 m/s with a downward direction?
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