An Olympic runner leaps over a hurdle. If the runner’s initial vertical speed is 2.2 m/s, how much will the ru

An Olympic runner leaps over a hurdle. If the runner's initial vertical speed is 2.2 m/s, how much will the runner's center of mass be raised during the jump?

4 Answers

  • The equation for vertical velocity is V=Vi-gt. The moment of peak altitude is when Vi-gt=0

    t=Vi/g. The vertical distance traveled to the peak altitude is then d=Vit-.5gt^2 Substitute in value for t

    d=[Vi*^2]/g-.5g[Vi/g]^2=.5[Vi^2]/g

    =.247m

  • in this type of Questions the acceleration of gravity is assumed constant because the altitude of jump is short.the basic differential equation of linear movement (adx=vdv).since the a=g=constant you will reach the equation :

    2a(x-x0)=v^2-v0^2.

    the movement of the runner in Cartesian coordination is divided between horizontal and vertical motion,as far as you need the ultimate latitude,we just consider the vertical linear motion,and the data would be:

    x0=0

    a=-g(approximately: -10)

    v0=2.2

    v=0

    by substituting the data in the equation:

    the answer (x) would be: 0.242 (m) (presuming that g=10)

  • The simplest way to do this is with potential and kinetic energy.

    You know through conservation of energy, that the initial kinetic energy becomes potential energy at the top of the jump, therefore:

    E = m*v*v/2 = mgh, cancel the m's

    v*v/2 =gh

    h = v*v/2g = 0.247 m

  • projectile problem. since you know the initial (vertical) velocity, acceleration (gravity), final (vertical) velocity, and starting position (0m), you can use vf^2=vi^2 + 2ad to find the final position.

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