Dynamics_Part10

Dynamics_Part10 - Problem Set 2: Problem 1. Problem: A ski...

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Unformatted text preview: Problem Set 2: Problem 1. Problem: A ski jumper has an initial velocity v = V i and follows the indicated trajectory, landing on the inclined slope at time a distance from the beginning of the slope as shown in the figure. The slope is inclined to the horizontal at an angle . You can ignore effects of friction on the ski jumper's motion. (a) Determine and as functions of V , g and , where g is gravitational acceleration. (b) At what angle , relative to the slope, is the ski jumper's velocity vector when he lands? Express your answer as a function of . (c) For what slope angle is a maximum, and what is the maximum value of ? HINT: For any function u(), du/d d tan-1 u = d 1 + u2 z .. ... . ....... ..... .......... ............. .......... ................ ................... ................ .................... . ........................ ...................... ........................... ............................. ........................... ............................... ................................... ................................. ..................................... ......................................... ....................................... .......................................... . .............................................. ............................................ ................................................. .................................................... ................................................. ..................................................... ......................................................... 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................................................................................... ................................................................................ .................................................................................. ...................................................................................... ....................................................................................... .............. ...................................... . ................................................................................... . ...................................................................................... ...................................................................................... . . .. ... . . . . . . . .. . . . . . . . . .. .. ................................................................................ .................................................................................... ................................................................................. . . ................................................................................... .................................................................. ................................................................................... .............................................................................. ................................................................................. . ... . . . ... ................................................................................. .. . .............................................................................. ............................................................................. ........................................................................... . ... . .. . ......................................................................... .......................................................... ....................................................................... ..................................................................... .................................................................. ................................................................ ............................................................ ............................................................ .......................................................... . . . .. .. . . . . . . .. ........................................................ ................................................... ...................................................... .... ... ... . . .. .. ... ... ..... ................................................. . . ... .... .. .. .. ... . ................................................ ............................................. ............................................ .......................................... . . . . . . ... .... ........................................ ................................... ..................................... .... . .. .. ... ... . ............................. ................................ .............................. ........................... ......................... . .. ... ..... ....................... ... .... ................... .................... 2 ................ .............. ............ .......... ........ .. .. . ..... v=V i x g = -g k So ut on: Because we gnore fr c on he mo on of he sk umper s sub ec on y o he cons an grav aona acce era on Deno ng me by t and e ng t = 0 when he sk umper aunches from he op of he s ope he equa ons and n a cond ons govern ng he mo on are d x =0 dt2 x(0) = 0 and x(0) = V d2 z = -g z(0) = 0 and z(0) = 0 dt2 (a) In egra ng w ce and mpos ng he n a cond ons he so u on o he equa ons for x and z s x(t) = V t When he sk umper ands a and 1 z(t) = - gt2 2 me we know ha x( ) = cos and z( ) = - s n = cos V 2 Therefore we have V = cos and 1 1 - s n = - g 2 = - g 2 2 cos V = ...
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This note was uploaded on 05/12/2010 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.

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