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What is the maximum height reached by the arrow?

  1. Apr 29, 2004 #1
    An archer shoots an arrow from a height of 5 ft. at an angle of inclination of 30 degrees with a velocity of 300 ft/sec. Write the parametric equations for the path of the projectile and sketch the graph of the parametric equations. If the arrow strikes a target at a height of 5 ft then how far is the target from the archer? For how many seconds is the arrows in flights. What is the maximum height reached by the arrow?
  2. jcsd
  3. Apr 30, 2004 #2


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    I can think of a number of ways to do this, some involving solving differential equations. Since you have not shown any work of your own on this problem or even what you feel you DO understand about it, I have no idea which method would be appropriate for your level and cannot answer.
  4. Apr 30, 2004 #3
    You don't need differential equations to solve this.

    Look, the equations of ballistic motion without air resistance are:

    x[t]=x0+v0*t*cos y[t]=y0+v0*t*sin+a*(t^2)

    where (x0,y0) is initial point, v=initial velocity, b=angle with x-axis, g=acceleration due to gravity, and t=time since launch. Therefore, the parametric equations are

    x[t]=300*cos[30]*t y[t]=5+150*t+(-32)*(t^2)

    It starts with intial vertical velocity v0=150 and hits the target with vertical velocity -150. Since v=v0+a*t, we have -150=150-32*t, giving a total flight time of t1=9.38 seconds.

    The range is given by plugging t1 into the formula x=x0+v0'*t+a'*(t^2)/2 along with x0=0, v0'=v0*cos=300*cos[30], a'=horizontal acceleration=0.


    Which should give you the right answer when put into a calculator.
  5. Apr 30, 2004 #4
    Forgot height. Sorry! The maximum or minimum of a parabola y=ax^2+bx+c occurs at x=-b/(2a), so the max height occurs at t=150/64. Plug that into y[t] to get max height.
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