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Projectile Motion Question

  1. Nov 10, 2007 #1
    I am not completely sure how to attempt this problem but I have given it a shot. If anyone knows how to do this please look it over, or post your attempt. Thank you!
    1. The problem statement, all variables and given/known data
    An object was shot at an angle of 49 degrees and landed 60m away at the same height. The initial velocity was 23m/s.
    WHat is:
    The length of time that the object was in the air
    The maximum height it reached

    2. Relevant equations


    3. The attempt at a solution









    so the maximum height would be 26m at 2.3s?
  2. jcsd
  3. Nov 10, 2007 #2
    You have to split the initial velocity into its vertical and horizontal components.

    You acknowledged that but didn't really do it.

    Velocity vector is 23m/s 49degrees above the horizontal.

    Vertical component is Vsintheta, and horizontal is Vcostheta.
  4. Nov 10, 2007 #3
    Ok, here is what you should do. Since the object is shot at an angle, you have to consider the vertical and horizontal components of the initial velocity. Once you get that, you can find the time taken by using the equation: t=deltax/Vix
    Note that you could even find the time using vertical portion of the motion--it would be the same. Time is the same in such a case. After that since you have the time, acceleration and Viy, you can find the max height. Any help?
  5. Nov 10, 2007 #4
    Thanks for your replies they have helped a lot.
    I was just wondering... would the time at maximum height be half of the total time or not necessarily? I know that the distance at maximum height would be half of the horizontal distance.
    Also, for the vertical the initial velocity would be positive while the final velocity would be the same number but negative right?
  6. Nov 10, 2007 #5
    I made a mistake, I am sorry. Look you found the time the object was in the air, not the time when the object is at a max height. So you cannot use that time.
    So you have Viy and a. Now lets take an example. When you throw a ball, it goes up to a max height and stops for a some time, and drops back down. Similar in this case. Hence the max height will be when Vfy=0. So you have Vfy, Viy and a. Solve for Delta Y. That's you mas height..:)
  7. Nov 10, 2007 #6
    ok.. so the time that the object was in the air is the length of the whole event. Therefore, I would only use that time only to determine the velocity of the horizontal correct? (v=d/t) After this velocity is calculated, then I would use it and half of the horizontal distance to calculate the time at max height. Then I would substitute this time in for "t" for the vertical and use the appropriate acceleration formula to calculate the vertical distance?
  8. Nov 10, 2007 #7
    horizontal velocity (and vertical) can be found usint trinometry (by the components method that i mentioned earlier)
  9. Nov 10, 2007 #8
    Using the angle mentioned in the question and the initial velocity which is the hypotenuse, I would do this and setup a right angled triangle?
    and of course (Sine, Cosine, Tangent)
  10. Nov 10, 2007 #9
    In this case it would because it is a parabola. the objects returns to the point where it was launched.

    you can find the total time of flight by:

    x(horizontal displacement)=Vox(initial x velocity)*t

    then you can use half the time to find delta Y. Many ways to do it.

    eDIT: sorry two people helping at once isnt effective because we are showing you different methods. pinkyjoshi you can continue.
    Last edited: Nov 10, 2007
  11. Nov 10, 2007 #10
    No problem, I appreciate your efforts and time trying to help me :)

    So the horizontal, vertical velocity calculated using the trigonometry method in the previous post and in the method that I quoted above, both methods should yield the same answer right?
  12. Nov 10, 2007 #11
    Umm...my question is, for the 1st part, how would you find time, if you don't find the horizonal velocity by the trig method. I mean if you use V=d/t, for finding t we need V and t. Do you get what I am trying to say?
  13. Nov 10, 2007 #12
    Yes sorry i didn't clarify that..
    Since both the vertical and horizontal have times that are the same, then I used acceleration formula (v=initial+at) manipulated, inserted what I knew and I got the time. Then, since this time is the same as the horizontal time, then I used it and the distance given in the question to determine the horizontal velocity. (v=d/t)
    Is this correct?
  14. Nov 10, 2007 #13
    Um..for your 1st equation: v=initial+at
    Wht will your V be?
    remember that this is for the vertical part of the motion. So you know the initial v(Viy by trig.) and a. But you do not know V ant T
  15. Nov 10, 2007 #14
    the question says that the initial velocity is 23m/s. So wouldn't the final velocity be
    -23m/s? That is what I used as the vertical final velocity.
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