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Projectile and ramp problem

  1. Sep 19, 2010 #1
    1. The problem statement, all variables and given/known data
    A projectile starts from rest and moves 4.7 m down a frictionless ramp inclined at 21 degrees with the horizontal. the acceleration due to gravity is 9.8 m/s^2

    what what speed will it leave the ramp ?

    what will the range of the projectile if the bottom of the ramp is 2.1 m above the ground


    2. Relevant equations

    V=Vi + at
    delta X = Vi*t+1/2 a*t^2
    Vf^2 - Vi^2 = 2*a*deltaX

    3. The attempt at a solution

    Vertical,
    Viy=0
    ay=-9.8
    Vfy= - Vf sin 21
    delta y= -4.7 sin 21

    horizontal,
    Vix= ( i thought it to be zero at first but it cant be zero because then the delta x comes out to be zero. So im confused at this one )
    ax=0
    Vfx=Vf cos 21
    delta x= 4.7 cos 21

    so i used the third law and set it up as follows,

    Vfy^2 = 2 (-9.8) (4.7 sin 21)

    then the idea is to set the answer = Vf sin 21 and solve for Vf

    it comes out to be 16.0329

    but this is unfortunately not the right answer. can anyone help ?
     
  2. jcsd
  3. Sep 20, 2010 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi physics_noob2,

    I don't believe this is true; while it is on the ramp, there is another force besides gravity acting on the object.

    More importantly, while it is on the ramp it is only moving in a single direction, so this part of the problem only has one-dimensional motion. What is the acceleration in the direction of motion? Once you have that, you can find the velocity at the end of the ramp.


     
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