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Find Initial Velocity - Projectile Motion

  1. Sep 20, 2015 #1
    • Missing template due to originally being posted in technical forum.
    I've been trying to come up with the answer but it hasn't gotten me anywhere.
    I know that I need to find the angle to get the velocity.

    Q: A rock is projected from the edge of a 30.48 m tall building at some unknown angle above the horizontal. 5 seconds after, the rock strikes the ground at a distance of 48.76 m from the building. Determine the speed of the rock.

    delta X = 48.76 m
    delta Y = -30.48 m
    t = 5 s
    Angle (Not necessary)
  2. jcsd
  3. Sep 20, 2015 #2
    What did you try? It helps to see where you're making mistakes in your work.
  4. Sep 20, 2015 #3
    My teacher hasn't really taught us how to do it but gave us 5 motion equations:

    a = V2 - V1 / delta t
    delta d = 1/2 (V1 + V2) delta t
    delta d =V1*t + 0.5t^2
    delta d = v2*t - 0.5*a*t^2
    v2^2 = V1^2 + 2a*delta d

    I tried to find the angle using
    theta = tan-1(Vyf/Vxf)
    theta = tan-1(dy/dx)
    It gave me 32 degrees

    Tell me if this is right and where do I go from here?
    Last edited: Sep 20, 2015
  5. Sep 20, 2015 #4
    How did you get the velocities for that angle formula?
  6. Sep 20, 2015 #5
    I didn't since I used the distances:

    theta = tan-1 (30.48m/48.76m)
    =32 degree
  7. Sep 20, 2015 #6
    But that's not telling you what angle it's thrown at--it's telling you the angle between the horizontal and the landing point.
  8. Sep 20, 2015 #7
    Previously -
    Vix = V cos 32
    Viy = V sin 32

    I tried this:
    delta Y = Viy*t +0.5 * -9.8 * 5^2
    Viy = (delta y - 0.5*9.8*-t^2)/t
    Viy = (30.48 -122.5)/5
    Viy = -18.404 m/s
    Last edited: Sep 20, 2015
  9. Sep 20, 2015 #8
    It is very basic question in physics. If you know motion Graph,I think it is very easy.You do not need to know the angel of the rock to find out its time to fall on ground.your method is also wrong.you should draw the diagram.
  10. Sep 20, 2015 #9
    I hope you can solve this problem. If you cannot, than I will help you.
  11. Sep 20, 2015 #10
    So basically I have to use simple equations to solve for this such as

    V= d/t
    V = 48.76 m / 5s
    V = 9.75 m/s

    If this is not right, then can you please help me?
  12. Sep 20, 2015 #11
    it is also wrong
  13. Sep 20, 2015 #12
    I've never had a problem that is to find the initial velocity of a projectile motion and with a height at that.

    Let me try it out though.
  14. Sep 20, 2015 #13
    what you study and which level you have? do not mind because it is easy for me to explain you.
  15. Sep 20, 2015 #14
    In grade 12 and I don't have any previous knowledge on projectile motion. We haven't really covered it much in class.
  16. Sep 20, 2015 #15
    yes, you read the book ...........Physics for scientist and Engineers by tipler...............Hope you will get the answer.
  17. Sep 20, 2015 #16
    I need a faster approach to solving the problem. If you know how to solve it can you show me how?
    If not then, I guess I can get the book.
  18. Sep 20, 2015 #17
    Yes, that is your horizontal velocity. Notice that the total horizontal distance is 48.76 m, and it is travelled in 5 s. You can do this because the x and y velocities are completely independent of each other.

    Now you need to find the y component of the velocity and use both of them to find the total speed.

    The formula you used before is correct, but you solved incorrectly. You have the wrong sign for the 0.5*9.8 t^2 term on the left side. Acceleration is negative, so you should be adding it to the left side. It should be [itex]-30.48 + \frac{1}{2} 9.8 (5^2)[/itex]. Please do go through each step one by one to see that this is the case.
    Last edited: Sep 20, 2015
  19. Sep 20, 2015 #18
    Yes, but please let Calaereth attempt it as they are the one who needs help.
    Last edited by a moderator: Sep 21, 2015
  20. Sep 21, 2015 #19
    Thank you very much. This has helped me a lot. :)
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