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Homework Help: Another projectile question

  1. Mar 8, 2005 #1
    God I hate projectiles..

    A PHYS1001 student uses a slingshot to project a pebble at her lecturer
    (shoulder height) who is standing 40 m away. After extensive experiments she
    finds that to hit the target she must aim 4.85 m above the target. Determine the
    velocity of the pebble on leaving the slingshot and the time of flight.

    Can some one start me off, with the correct formula to use. Im stumped..
  2. jcsd
  3. Mar 8, 2005 #2


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    Science Advisor

    The projectile is required to hit the target at the same height it leaves the student. Furthermore, the Angle θ (above horizontal) at which the projectile leaves the student is given by:
    tan(θ) = (4.85)/(40) = (0.121) ::: ⇒ θ = (6.91 deg)

    The standard equations for projectile height z(t) under constant gravitational acceleration "g" and for horizontal distance "x" are given by:
    z(t) = z0 + vz0t - (1/2)gt2 :::: Eq #1
    x(t) = x0 + vx0t :::: Eq #2

    Let T be the time of flight. The problem requires that {z(T) = z0} and that {x(T) - x0 = (40 m)}. Placing these into Eqs #1 & #2 we get:
    z(T) = z0 = z0 + vz0T - (1/2)(9.81)T2
    ⇒ 0 = vz0T - (1/2)(9.81)T2
    ⇒ 0 = vz0 - (4.91)T
    ⇒ T = vz0/(4.91) :::: Eq #3

    x(T) - x0 = 40 = vx0T
    ⇒ T = (40)/vx0 :::: Eq #4

    We now multiply Eq #3 by Eq #4:
    T2 = {40/4.91}*{vz0/vx0} =
    = (8.15)*tan(θ) =
    = (8.15)*(0.121) =
    = (0.986)
    ⇒ T = (0.993 sec)

    Placing this last result into Eq #4 and solving for vx0:
    (0.993) = (40)/vx0 ::: ⇒ vx0 = (40)/(0.993) = (40.3 m/sec)
    ⇒ v0 = vx0/cos(θ) = (40.3)/cos(6.91 deg)
    ⇒ v0 = (40.6 m/sec)

  4. Mar 8, 2005 #3
    life saver!
  5. Mar 8, 2005 #4
    There's another way, though not too different from xanthym's.

    The trajectory that the pebble follows is a parabola under the (assumed constant) gravitational field of earth. If the pebble were projected with an initial velocity [itex]\vec{u}[/itex] at an initial elevation [itex]\alpha[/itex] (computed just as xanthym did) then the equation of the trajectory turns out to be,

    [tex]y = xtan\alpha - \frac{1}{2}\frac{gx^2}{u^2}sec^2\alpha[/tex]

    In principle you can find u directly from this equation given y = 0, x (= range) and alpha. Once you have that, use Range = u\cos\alpha*Time of Flight to find the time of flight.

    The equation here isn't really a shortcut as in some cases, the algebraic calculations are easier if you follow the step-by-step fundamental method. Yet its worthwhile to know the equation, as it relates all the parameters of motion and gives you y as a function of x.


    PS--It would be good exercise for you to derive this equation of trajectory yourself. Hint: xanthym's equations are sufficient to do this.
  6. Mar 9, 2005 #5
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