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Projectile motion

  1. Nov 1, 2003 #1
    A projectile is fired with an intial speed of 113 m/s at an angle of 60 degrees above the horizonal from the top of a cliff 49.0 m high

    a) time to reach the max height
    b) max height above the base
    c) total time in the air
    d) horizonal range of the projectile

    I used the formula t^2 = Vo*sin 60/g
    t^2 = 113*sin60/9.80 = 9.98 seconds
    And after that i am totally lost and can't find any real go examples in the book.

    where can i go from here and what formulas can i used Still going over notes and reviewing the book.

    I also drew a picture
     
  2. jcsd
  3. Nov 1, 2003 #2

    HallsofIvy

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    Staff Emeritus
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    The moral to this story is "don't use formulas if you don't know where the came from".

    The acceleration is -g vertically. There is no acceleration horizontally.

    Since -g is a constant, in time t, the vertical speed will change by -gt. The horizontal speed will not change.
    Since the initial vertical speed was 113 sin(60), that vertical speed at time t will be 113 sin(60)- gt. It will continue going up until that speed is 0: the maximum height will occur when 113 sin(30)- gt= 0 or t= 113 sin(60)/g.

    Assuming your "t^2 = 113*sin60/9.80 = 9.98 seconds" was intended to
    solve (a) did you really get lost here? I would think that taking the square root of t^2 would be an obvious way to get t!

    (b) asked for the maximum height"above the base"
    You know the vertical speed is 113 sin(60)- gt. Integrating that gives h(t)= 113 sin(60)t- (1/2)gt^2+ C for the height at time t. When t= 0, h(0)= C. What was the height of the projectile ABOVE THE BASE OF THE CLIFF when it was fired? Since you know, from (a) the time when it was at its highest point, put that value of t into h(t).

    (c) asked for "total time in the air". It won't be in the air any more when it hits the ground! Since it was launched away from the cliff, that will be when h(t)= 0. Solve the equation you got in (b) to find t when h(t)= 0.

    (d) asked for horizontal range
    The initial horizontal speed was 113 cos(60) and there was no acceleration. The horizontal distance moved is 113 cos(60) t.
    Put in the t from (c) (when the projectile hit the ground) to find the horizontal distance in that time.
     
  4. Nov 1, 2003 #3
    How about this one:


    R= ((Vo^2sin2(angle)))/g

    Nautica
     
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