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Homework Help: Projectile motion

  1. Oct 22, 2007 #1
    1. The problem statement, all variables and given/known data
    A high jumper whose center of gravity is 1.1m above the ground h been clearing 1.8m using a western roll in which the take off velocity is at an angle 60deg with the horizontal. With what speed must be take off? how far back from the bar must he take off?

    I understand the question except for the 1.1m above ground center of gravity.
    What significance does this have?

    2. Relevant equations

    3. The attempt at a solution
    I cant proceed until I understand what the center of gravity is about.
  2. jcsd
  3. Oct 22, 2007 #2
    His centre of gravity must move up and over the bar to make the jump
  4. Oct 22, 2007 #3
    Would it be alright if i subtracted 1.1m from 1.8m and then began my calculations?
  5. Oct 22, 2007 #4
    Yes, that's fine
  6. Oct 22, 2007 #5
    0=(VoSin60)^2 + 2(-9.81)(0.7)
    Vo = 4.279m/s -- is the initial velocity he must take off with.

    0.7 = -0.5(-9.81)t^2
    t= 0.3778secs -- is the time it takes to reach the highest point i.e. the bar

    x=0.808m -- the distance away from the bar he must jump

    looks reasonable, can someone concur?

    another question i have is, if an object is directly projected downwards at an angle with the horizontal, which is the angle of depression.

    Do i use this angle for Vox and Voy or 90 minus that angle?
    Last edited: Oct 22, 2007
  7. Oct 22, 2007 #6
    is that correct or do i have to minus by 90?
  8. Oct 22, 2007 #7
    Your calculations look correct to me.

    You shouldn't liberally apply cosines and sines without thinking about what you're doing. Remember, the whole point of using the trig functions is to resolve a vector into mutually perpendicular axes, in this case, to resolve the velocity into the x- and y-directions. Draw a diagram with the velocity and the angle, then axes representing the x- and y-axes, and see what trig functions you need to resolve it into a given direction.
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