1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Velocity in the horizontal and vertical direction

  1. Jan 1, 2010 #1
    1. The problem statement, all variables and given/known data

    Santa is in trouble. His reindeer just made a rough crosswind landing on the snow covered apex of this gingerbread house. Upon impact the hitch breaks free and the reindeer shoot skyward leaving Santa teetering on the very peak. As he slightly shifts the sleigh tilts forward. Remembering his physics, Santa quickly surveys his predicament. He weighs 256 pounds, his center of mass is located 4 feet vertically above the roof, his skis are waxed, the roof has a 1-in-5 pitch, the side of the house is 12 feet high, the width of the house is 40 feet from the left edge to the right roof edge and the snow on the ground is 5 feet deep.

    Determine Santa's velocity in both the horizontal and vertical direction when he hits the snow. (Hint: make sure you use the vertical distance from the center of mass of santa to the top of the snow)

    in previous questions i found the forward velocity as soon as he leaves the roof (Vx)= 17.53 ft/sec
    and the downward velocity as soon as he leaves the roof (Vy)= 3.51 ft/sec

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Jan 1, 2010
  2. jcsd
  3. Jan 1, 2010 #2
    The angle the roof makes with the horizontal is tan A=1/5 or about 11 degrees. The distance from the roof peak to the side wall is 20 feet (40/2). Because the roof pitch is 1:5 the height of the peak above the side wall is 4 feet. Santa's center of mass is 4 feet above the peak. Santa is going to fall 8 feet before he slides off the roof. Using energy methods his PE is converted to KE. His velocity as the roof is v=sqrt(2gh). You know the roof angle so you can find the Vx and Vy components of V.

    Now you have a projectile problem with Santa falling 12 feet in the negative Y direction (assuming the snow is at Y=0. You know Y=Vyt-.5g*T^2. You know Y and Vy (watch signs) solve for t. The new Vy, when he hits the snow) is Vy=Vy0-g*t. Vx doesn't change.
  4. Jan 1, 2010 #3
    i thought of using the equation for time as: dy= Viyt + (ayt^2)/2
    with ay: 32 ft/sec^2

    i just didnt know what to put for distance in the y. the snow is 5 feet deep so do i subtract 5 from the 12 feet? or do i add 4 feet to the 12 feet and then subtract 5?

    and then once i got the time i could solve for the velocity in the x and y with these formulas:
    Vfx@___sec = Vix + axt=
    Vfy@___sec= Viy+ayt=
    Last edited: Jan 1, 2010
  5. Jan 2, 2010 #4
    Part 1: Santa's center of gravity falls 8 feet to the edge of the roof. Use energy methods to find Vx at Vy at the edge o fthe roof.

    Part 2: Now Santa is going to fall 7 feet to the snow (12-5) so that is Y.

    Y=Vy*t-.5g*t^2 where Vy is found above, solve for t

    To find the new Vy(final)=Vy(original)-g*t this is the velocity in the Y direction when santa hits the snow. Vx is the same as at the edge of the roof.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook