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Homework Help: Tough problem (for me at least )

  1. Nov 10, 2004 #1
    Alright, so here's the problem:

    So I didn't want to explain everything I tried, so I just scanned my paper ;). Here it is:


    Can anyone see what I'm doing wrong? The correct answer is supposed to be 10.9 m/s...
  2. jcsd
  3. Nov 10, 2004 #2
    Don't use forces, use energy

    the velocity whe the skier leaves the cliff is...
    [tex]\frac{1}{2}mv^2 = mgh[/tex] while h = 10.4m sin 25
    v = sqrt(2g(10.4 sin 25)) = 9.29 m/s

    the skiers velocity is 25 degrees below the horizontal, thus
    [tex]v_x = v cos 25 \ v_y = v sin 25 [/tex]
    thus solve for the time to fall the height 3.5 m
    [tex] 0m = 3.5 m - v sin 25 t - 1/2 g t^2[/tex]
    t= 0.53s

    and when the skier makes impact the velocity is
    [tex]v_y = v sin 25 + 9.8 (0.53 s)[/tex]
    [tex]v_x = v cos 25[/tex]
    so the total velocity at impact is..
    [tex] v = sqrt( (v_x)^2+(v_y)^2) = 12.4128 m/s[/tex]

    so i believe your answers wrong doing this is also very complicated but i thought would make u believe the answer more cause I use projectiles, gravity is a conservative force so therefore it doesn't matter what path the skier takes to drop the height u can find the velocity by 1/2mv^2 = mgh
    where h is (10.4m sin 25 + 3.5m) the total height from the ground. you'll get the same result.

    edit: i forgot the coff of friction, however use work engery principle in the same way i did this problem that might help
    Last edited: Nov 10, 2004
  4. Nov 10, 2004 #3
    Your sum of all x component forces :
    Your sum of all y component forces :
    [tex]\Sigma F_{y}=ma_y[/tex]
    Last edited: Nov 10, 2004
  5. Nov 10, 2004 #4
    Thanks guys, really appreciate your help.
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