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Projectile Motion Q

  1. Dec 9, 2005 #1
    A diver leaves the springboard with a vertical velocity of 5.7m/s, and horizontal velocity of 1.2 m/s, and a height of 3.49m. How long will it take her to return to the same height? What horizontal range will she cover in this time?

    Would I use: Time = 2Vvert/g for the first part? And Range = (Vhor)(Time) for the second part?
    1st part --> Time = 1.16 s
    2nd part --> Range = 1.4 m

    Is this correct? :confused:
  2. jcsd
  3. Dec 9, 2005 #2
    In order to solve this qn, perphaps using the formulae for 2D motion will be the easiest.

    Since time is the quantity tat u r finding..of course it must be in ur formula.. Given the height to be the 3.49m... for the diver to reach back the same displacement later in the dive, u will need to use the formula [tex] s=ut+1/2gt^2 [/tex] to solve.. the formula for quadratic eqns will get u 2 values for t.. the ans will be the later t.
  4. Dec 9, 2005 #3
    oh ya.. for the previous post, initial velocity, u, is the initial vertical velocity..

    For the 2nd part.. assuming zero air resistance, the horizontal velocity the diver travels will be the initial horizontal velocity times the time u found in (a)
  5. Dec 10, 2005 #4


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    In this case s=0 so the t cancels out and you get the equation
    t = 2Vy/g. No need to solve quadratic equations.
  6. Dec 11, 2005 #5
    Projectile Motion Q...cont.

    daniel_i_l: So my work that is shown in my initial post is correct then?
  7. Dec 12, 2005 #6


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    yes, it is
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