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Another projectile question

  1. Oct 11, 2003 #1
    I need a little help with another projectile question, id be very obliged if anybody could provide some:

    (A) A golf ball at rest on horizontal ground is struck so that it starts to move with velocity 3u I + u j ( “I” and “j” are unit vectors along and perpendicular to the ground) . In its flight, the ball rises to a max height of 15 m. Calculate:
    (i) The value of u
    (ii) The magnitude and direction of the velocity with which the ball strikes the ground.

    (i) What ive got is Sy = 15m, when Vy = 0, time when Vy = 0 is (u/g).
    I then subbed this into Sy= ut - (1/2)gt^2 = 15 and got u =

    (ii) Velocity, when Sy=0.
    Time when Sy=0 is (2u/g)

    I then subbed this into my Vx (3u) and Vy (u - gt) equations to get √2646i - √1176 j.

    Im wondering wheter this answer is right, it just doesnt sound right to me, the negative j-value maybe is a bit off?

    I then got the Tan of (-√294 / √2646) = -1/3 = -18degrees 26 minutes.
    I interpreted this as being E18.26.S, which im not too sure about and is a partial guess.

    Part (B) is where im really getting stuck:

    A particle is projected from a point p with an initial speed 15m/s, down a plane inclined at an angle 30 degrees to the horizontal. The direction of projection is at right angles t the inclined plane. (The plane of projection is vertical and contains the line of greatest slope), Find:
    (i) the perpindicular height of the particle above the plane after t seconds and hence, or otherwise, show that the vertical height h of the particle above the plane after t seconds is 10√3t - 4.9t^2.
    (ii) The greatest vertical heigt it attains above the plane (the max value of h) correct to two places of decimals.

    Ive figured from my sketch that the perpindicular height above the plane is (Sy/cos 30). Im not too sure about how i should go about getting the Ux and Uy though, since the particle is fired at right angles to the plane.

    from the question i know that Uy is 10-√3, but im assuming ive got to prove that also. Thats about all ive gotten so far.

    I'd appreciate any help, and sorry about the crappy diagram...
    Last edited by a moderator: Feb 6, 2013
  2. jcsd
  3. Oct 12, 2003 #2
    Any help would be very much appreciated...
  4. Oct 12, 2003 #3


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    It would have helped if you had stated that "Sy" was the height of the ball and Vy the y coordinate of speed instead of just using the variables without definition.

    Yes, that is correct.

    Your y-value looks to be twice the correct value. Since t= 2u/g,
    u- gt= u- g(2u/g)= -u.

    When the projectile hits the ground its x-component of velocity is still 3u but its y component is -y (because of the symmetry of the trajectory). Of course, tan(θ)= -u/3u= -1/3. I get θ= 18 degres 24 minutes approx.

    Yeah, I can see how this would be hard. The way I would do this would be to treat the particle as a regular projectile following a hyperbolic trajectory. Since the inclined plane is 30 degrees down from the horizontal, the initial velocity of the particle is 60 degrees up (so that 60+30=90 is the angle between the plane and the initial speed). The intitial velocity of the particle is given by
    Vx= 15 cos(60)= (1/2)15= 7.5 m/s, Vy= 15 sin(60)= 7.5 √(3) m/s.
    The position (measured from the initial point) at any t is Sx= 7.5t and Sy= 7.5 √(3) t- 4.9 t2.

    Now look at the inclined plane. Since it is inclined 30 degrees downward its slope is tan(-30)= -1/√(3). Its equation is
    y= (-1/√(3))x.

    The height of the partical above the plane is "height of partical - height of plane"= 7.5 √(3)t- 4.9t2+ (1/√(3))(7.5t)
  5. Oct 12, 2003 #4
    Thanks for the help
    Sorry about not defining Sx, Vx, etc, im just so used to using them all the time i just assumed everybody else was too
    Thanks again
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